$f: \mathbb{R^+} \to \mathbb{R^+}$ which satisfies $f(xf(y))=x^2y^3.$
\begin{align} & \text{let } P(x, y): f(xf(y))=x^2y^3. \\ \ \\ P(1, x): \; & f(f(x))=x^3. \\ \ \\ P(f(x), y): \; & f(f(x)f(y))=f(x)^2y^3. \\ \therefore \; & f(f(x)f(y))=x^3f(y)^2. \\ \Rightarrow \; & f(x)^2y^3=x^3f(y)^2. \\ & \dfrac{f(x)^2}{x^3}=\dfrac{f(y)^2}{y^3}. \\ \therefore \; & f(x)=kx^{\frac {3} {2}}. \\ \ \\ &f(f(x))=k\left(kx^{\frac {3} {2}}\right)^{\frac {3} {2}}=k^{\frac {5} {2}} x^3 = x^3. \\ \therefore \; & k=1, f(x)=x^{\frac {3} {2}}. \\ \ \\ \end{align}
But, putting this to the original FE, we get the contradiction. Did I prove that there doesn't exist any function $f$ that satisfies the above? If not, did I miss something?
Linked question: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(yf(x))=x^2y^4.$
Since the verification was done at the comment, I'll put this as an answer with the community wiki.