If a function can only be defined implicitly does it have to be multivalued?

Not necessarily. Consider the graph $G$ in ${\mathbb R}^2$ of the points $(x,y)$ such that $$ y^5+16y-32x^3+32x=0. $$ This example comes from the nice book "The implicit function theorem" by Krantz and Parks.

Note that this is the graph of a function: Fix $x$, and let $F(y)=y^5+16y-32x^3+32x$. Then $F'(y)=5y^4+16>0$ so $F$ is strictly increasing. Since $F(y)\to-\infty$ as $y\to-\infty$ and $F(y)\to\infty$ as $y\to\infty$, the equation $F(y)=0$ has at least one solution (by the intermediate value theorem), but then it has precisely one. This means that the graph $G$ describes a function (in the strict, traditional sense, not a multivalued function).

However, there is no formula (in the traditional sense) giving us $y$ explicitly as a function of $x$.

This specific example shows one of the reasons why this may happen. Here, we have $y$ as a solution of a quintic equation, but in general there is no algebraic formula that gives us these solutions; this is a deep result of Abel and Galois.

There are similar results stating that certain equations do not have elementary solutions (for example, certain differential equations), so they may define a function but we may not have a "formula" for it. In fact, the concept of what a "function" is underwent a few revisions before reaching its current modern form, precisely as we realized the need for versions more general than "given by an explicit formula". A nice account of the evolution of the term is in the paper by Penelope Maddy, "Does $V$ equal $L$?", J. Symbolic Logic 58 (1993), no. 1, 15–41. jstor, projecteuclid

The notion of an "implicitly defined function" (IDF for short) has nothing to do with the question whether some such function can be expressed "in finite terms". The general reason for the appearance of an IDF is that we are given a relation $F(x,y)=0$ between two variables $x$ and $y$ (these may be vector-valued), where a priori these two variables have "equal rights". E.g., the relation $F(\cdot,\cdot)$ may be a natural law that connects the temperature $T$ and the pressure $p$ of some amount of gas in a closed vessel.

In such a case the question arises whether, given the value of $x$, the relation $F(x,y)=0$ determines a unique value $y$. If this is (maybe in a restricted sense, see below) the case, this value $y$ is a function of the chosen $x$; one is allowed to write $y=f(x)$ and says that the function $f$ is "implicitly defined" by the relation $F(x,y)=0$. If $F$ is given as an explicit expression in the variables $x$ and $y$ and one is lucky one may algebraically solve the equation $F(x,y)=0$ for $y$ and get an explicit expression for $f$. Usually one cannot solve for $y$, but one nevertheless is interested, e.g., whether $f$ is continuous, or one would like to know the value of $f'(x)$.

The theorem on implicit functions gives an answer to such questions. The essential point is that you need not to be able to solve your equation explicitly for variable $x$ and are nevertheless able to compute $f'(x)$ in individual points $x$. The theorem is a local theorem. It says that any point $(x_0,y_0)$ that satisfies the equation is the center of a rectangular window $W=U\times V$ such that the set $\{(x,y)\in W | F(x,y)=0\}$ can be viewed as the graph of a "local" function $f:\ U\to V$. There is a formula for $f'(x_0)$, namely $$f'(x_0)=-{F_x(x_0,y_0) \over F_y(x_0,y_0)} \ .$$ This formula shows that we forgot to mention the main assumption for all this to hold: One needs $F_y(x_0,y_0)\ne 0$.

Explicit functions are great for expressing a dependent variable as a function of an independent variable. But it's not always possible, and even if possible, not always most appropriate to arbitrarily define one variable in terms of another (or in terms of many).

Some implicitly defined functions (Ex 1), but not all (Ex 2), are not (strictly speaking) functions at all.

• Ex 1 (not a "strict" function): $f(x, y) = \{(x, y) | x^2 + y^2 = r^2, r>0\}$ where $r$ is some positive constant, defines a circle of radius $r$. It is not a function in the strict sense of the term, since, if $(x, y)$ is a solution, then so is $(x, -y)$. (Or alternatively, if $(x, y)$ is a solution, then so is $(-x, y)$). However, we can define each of the semicircles for $y \geq 0$ and for $y < 0$ as explicit functions, such that the graphs of both functions forma circle.

• Ex 2 (strict function expressed implicitly): There are some strict functions which are best expressed as implicit functions, or for which one variable cannot be expressed explicitly as a function of another variable using only elementary functions, like $$e^y = \frac {xy}{x+y}$$

Rather than elaborate here, Wikipedia does a nice job of clarifying the distinction between implicit and explicit functions, and/or when one form is preferable to the other:

I also encourage you to take a look at the Implicit Function Theorem:

In multivariable calculus, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.