a) Take $x=1$ then $f(yf(1))=f(y)$ with (1) gives $f(1)=1$ b) Take $y=1$ then $f(f(x))=x^2f(x)$ c) Take $x=y$ then $f(xf(x))=x^2f(x^2)$ d) Take $y=f(x)$ then $f(f(x)^2)=x^2f(xf(x))=x^4f(x^2)=f(f(x^2))$ so $f(x^2)=f(x)^2$ We have $f(xf(x))=x^2f(x^2)=x^2f(x)^2=(xf(x))^2$ Take $y=xf(x)$ then $f(y)=y^2$ and its indeed solution.

Unless I am mistaken, you have shown that $ f(y)=y^{2}$ for all $ y$ that are of the form $ xf(x)$ (please correct me if I am wrong). In order to show that $ f(y)=y^{2}$ for ANY $ y$, we may proceed as follows: Assume $ f(y)>y^{2}$. Then, $ f(f(y))>f(y^{2})=f(y)^{2}$, because of (1), and where d) has been used. Then, using b), $ f(y)^{2}<f(f(y))=y^{2}f(y)$, or $ f(y)<y^{2}$, absurd. Similarly, assume $ f(y)<y^{2}$, and in the same way find that $ f(y)>y^{2}$, absurd again. Or $ f(y)=y^{2}$. Substitution in the original equation yields $ x^{4}y^{2}$ at both sides, or $ f(y)=y^{2}$ is indeed the only solution.