A function $z=z(x,y)$ is given implicitly by the function $$f\left(\frac{x}{y},\frac{z}{x^{\lambda}}\right)=0$$ where $\lambda\in\mathbb{R},\lambda\neq0$.
I have to show that if $f(u,v)$ is differentiable and $$\frac{\partial f}{\partial v}(u,v)\neq0$$ then $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\lambda z$$
I tried to do this using the Implict function theorem: $$\frac{\partial z}{\partial x}=-\displaystyle\dfrac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$ So calling $$F(x,y)=f\left(\frac{x}{y},\frac{z}{x^{\lambda}}\right)$$ and applying the chain rule, I got: $$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial x}$$ so $$\frac{\partial F}{\partial x}=\frac{\partial f}{\partial x}\cdot\frac{1}{y}-\frac{\partial f}{\partial y}\cdot\frac{\lambda z}{x^{\lambda+1}}$$ But I don't know if is this the right way and, if it really is, what I'm supposed to do from now on?
I did all this trying to find some expression to $\frac{\partial f}{\partial x}$, but I don't know if it worked.