Is there a continuous $f(x,y)$ which is not of the form $f(x,y) = g_1(x) h_1(y) + \dots + g_n(x) h_n(y)$

Let's call $f$ an $n$-SOP if we can write

$$f(x,y) = \sum_{k = 1}^n g_k(x)\cdot h_k(y)$$

with continuous functions $g_k, h_k \colon [0,1] \to \mathbb{R}$. If $f$ is an $n$-SOP, for every family $x_1 < x_2 < \dotsc < x_r$ of $r > n$ points in $[0,1]$, the set

$$\left\{ \begin{pmatrix} f(x_1,y) \\ f(x_2,y) \\ \vdots \\ f(x_r,y)\end{pmatrix} : y \in [0,1]\right\}$$

is contained in an $n$-dimensional linear subspace of $\mathbb{R}^r$.

But

$$\begin{pmatrix} \exp (x_1\cdot 0/r) & \exp (x_1\cdot 1/r) & \cdots & \exp (x_1 \cdot (r-1)/r) \\ \exp (x_2 \cdot 0/r) & \exp (x_2 \cdot 1/r) & \cdots & \exp (x_2 \cdot (r-1)/r) \\ \vdots & \vdots & & \vdots \\ \exp (x_r\cdot 0/r) & \exp (x_r\cdot 1/r) & \cdots & \exp (x_r \cdot (r-1)/r)\end{pmatrix}$$

is a Vandermonde matrix, hence has rank $r$. Therefore $(x,y) \mapsto e^{xy}$ is not an $n$-SOP.

We have a writing $f(x,y) =\sum_{i=1}^n g_i(x) h_i(y)$, equivalently $$f(\cdot, y) = \sum_{i=1}^n h_i(y) g_i(\cdot)$$ means that the system of function $(f(\cdot, y))_y$ is contained in the span of n functions $g_i(\cdot)$, $i=1, \ldots, n$, equivalently, this system of functions $(f(\cdot, y))_y$ has rank at most $n$ ( remember: functions are "vectors" ).

But for a general function $f(x,y)$ ( analytic, not a polynomial) the system $(f(\cdot, y))_y$ is infinite dimensional ($e^{x y}$ (@Daniel Fischer: answer), $\frac{1}{x+y}$, $\sqrt{x+y}$, $\frac{1}{1-x y}$ $\&$ c.)

It's worth looking at $f(x,y)$ as an infinite matrix indexed by $x$, $y$. There exists such a writing if and only if the matrix has rank $\le n$, that is, if and only if every minor $$\det (f(x_i, y_j))_{i,j=1, n+1}=0$$