Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers

They are the only possible solution. Proof is as follows:

Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime. Substituting in the original equation we get $$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$ Solving for $d$: $$d=\frac{\left( y_0+x_0\right) \,z}{x_0\,y_0\,\left( z-1\right) }$$ Since $x_0$ and $y_0$ are co-prime, for $d$ to be an integer, $x_0\,y_0$ should divide $z$. Hence we require $$ z= k ~x_0 ~y_0$$ Substituting in the equation for $d$ and solving for $k$: $$ k=\frac{d}{d\,x_0\,y_0-y_0-x_0}$$ This shows that $d$ is a multiple of $k$. Let $$ d= \mu k$$. Then $$k=\frac{k\,\mu}{k\,\mu\,x_0\,y_0-y_0-x_0}$$ Solving for $k$: $$k=\frac{1}{x_0\,y_0}+\frac{1}{\mu\,y_0}+\frac{1}{\mu\,x_0}$$ This implies that $1 \le x_0 \le 3$, $1 \le y_0 \le 3$, $1 \le \mu\le 3$

It is possible to eliminate some of the 27 possible values since $k$ has to be an integers this will result in 12 possible values for $(x_0,y_0,\mu)$ and two of the solutions are repetitions giving the 10 solutions mentioned in the problem.

HINT : You may suppose that $1\le x\le y\le z.$ This will make it easier to find the solutions.

We may as well assume $x\le y\le z$ (and then count rearrangements of the variables as appropriate). The smallest variable, $x$, cannot be greater than $3$ (or else $1/x+1/y+1/z\lt1/3+1/3+1/3=1$), nor can it be equal to $1$ (or else $1/x+1/y+1/z=1+1/y+1/z\gt1$). So either $x=2$ or $x=3$.

If $x=3$, then $y=z=3$ as well (for the same reason as before), which gives the solution $(x,y,z)=(3,3,3)$.

If $x=2$, then $1/2+1/y+1/z=1$ implies

$${1\over2}={1\over y}+{1\over z}$$

Applying the inequality $y\le z$ to this equation, we see that $y$ must be greater than $2$ but cannot be greater than $4$, so $y=3$ or $y=4$. Each of these gives a solution, $(x,y,z)=(2,3,6)$ and $(2,4,4)$.

Counting rearrangements, we get the OP's $10$ solutions and no others.

Hint: Deduce that none of $x, y, z \in \mathbb{N}$ exceeds $7$. This can be done by mathlove's hint above.

Additionally, you can show that there is only one $(a, 2, 2)$-tuple and thus use $(a, 2, 3)$ to bound the solutions.