Proving $x\frac{\partial z}{\partial x}+2y\frac{\partial z}{\partial y}=nz$, for $z(x,y)=x^nf(\frac{y}{x^2})$. Where is my calculation wrong?

Question

Given the function $z(x,y)=x^nf(\frac{y}{x^2})$ where is $f$ a differentiable function. Prove that equation: $$x\frac{\partial z}{\partial x}+2y\frac{\partial z}{\partial y}=nz \tag{$\star$}$$

My attempt is:

$$\frac{\partial z}{\partial x}=x^{n-1}f\left(\frac{y}{x^2}\right)(n-y)\quad\rightarrow\quad x\frac{\partial z}{\partial x}=x^{n}f\left(\frac{y}{x^2}\right)(n-y) \tag1$$

$$\frac{\partial z}{\partial y}=x^{n-2}f\left(\frac{y}{x^2}\right)\quad\rightarrow\quad 2y\frac{\partial z}{\partial y}=2yx^{n-2}f\left(\frac{y}{x^2}\right) \tag2$$

But $(\star)$ not meet. Wrong is my, or wrong is the $(*)$? Maybe it should be $$x\frac{\partial z}{\partial x}+x^2y\frac{\partial z}{\partial y}=nz \tag{$\star$}$$

Answer 1

$$\begin{array}{l}\frac{\partial z}{\partial x}=nx^{n-1}f(\frac{y}{x^2})-2yx^{n-2}f'(\frac{y}{x^2}) \\ \frac{\partial z}{\partial y}=x^{n-2}f'(\frac{y}{x^2}) \end{array}$$ then easily (*) obtained.

Answer 2

hint

$$\frac{\partial z}{\partial x}=$$ $$nx^{n-1}f(\frac{y}{x^2})-x^nf'(\frac{y}{x^2})\frac{2y}{x^3}$$

$$\frac{\partial z}{\partial y}=x^{n-2}f'(\frac{y}{x^2})$$

thus

$$x\frac{\partial z}{\partial x}=nz-2yx^{n-2}f'(\frac{y}{x^2})$$

$$=nz-2y\frac{\partial z}{\partial y}$$