Putting y=0 we get that f(0)=0 Putting y=1 we get that xf(x)=f(x^2) Now we put x^2 and y^2 instead of x and y. By this we after reduction and using this what we get moment ago we will have that yf(x^{2}y)=xf(y^{2}x) If we use it to original equation we get that f(x^{2}y)=f(xy) for xy(x^2 - y^2) not equal to 0. Now we can put xy=1 to this and we get that f(x)=xf(1), so there exist some a that f(x)=ax. But every such function meets the condition of our problem, so the answer is f(x)=ax for some real a.

P(x,0) -> x^2f(0)=xf(0) -> f(0)=0 The rest is easy

let y=1 we have:$(x^{2} -1)f(x)=xf(x^{2} )-f(x)$ so $xf(x)=f(x^{2} )$ it is easy to find that f is an odd function(notice $x^{2}=(-x)^{2} $) we have: $\frac{f(x)}{x} =\frac{f(x^{2}) }{x^{2} } $ and $\lim_{x \to \infty} \sqrt[n]{a} =1(a\in R_{+ } )$ so $\frac{f(a)}{a} =\frac{f(1)}{1} =f(1)=t$ so f(x)=kx Q.E.F