Implicit differentiation of a two variables function

Question

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.

Answer 1

$$df=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv=0$$ where $$du=d(x/y)=\frac{1}{y}dx-\frac{x}{y^2}dy$$ $$dv=d(z/x^\lambda)=\frac{1}{x^\lambda}dz-\frac{\lambda z}{x^{\lambda+1}}dx$$ We know that $$dz=\frac{\partial{z}}{\partial x}dx+\frac{\partial{z}}{\partial y}dy$$ So $$dv=d(z/x^\lambda)=\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx$$ Now $df$ becomes $$df=\frac{\partial f}{\partial u}\bigg( \frac{1}{y}dx-\frac{x}{y^2}dy\bigg)+\frac{\partial f}{\partial v}\bigg(\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx \bigg)=0$$ This equation is always satisfied when $$\frac{1}{y}dx-\frac{x}{y^2}dy=0$$ $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}dy-\frac{\lambda z}{x^{\lambda+1}}dx=0$$ From the first condition it follows that $$dy=\frac{y}{x}dx$$ Replacing this into the second condition $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}dx-\frac{\lambda z}{x^{\lambda+1}}dx=0$$ $$\bigg(\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}\bigg)dx=0$$ $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}=0$$ $$x\frac{\partial{z}}{\partial x}+y\frac{\partial{z}}{\partial y}=\lambda z$$