INMO!

Substitute y=0 and we get $f({ x }^{ 2 })=xf(x)$ .

Let's say $f(x)=xg(x)$ , then we get

$g({ x }^{ 2 })=g(x)$ .

It follows that $g(x)=c$ and $f(x)=cx$

Plugging back into the original equation, the solutions for c are 0 and 1 so $f(x)=0$ or $f(x)=x$

Thus, we have 2 solutions.

Upon observation, $f(x) = ax$ , $a \in \mathbb{R}.$ Substituting this function into the above functional equation yields:

$a(x^2 + axy) = x(ax + ay) \Rightarrow ax^2 + a^2xy = ax^2 + axy \Rightarrow a^2 = a =\Rightarrow a = 0,1$ .

Hence, we have $f(x) = 0$ or $f(x) = x$ .

the ans is 2