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.

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    $\begingroup$ Some observations: (i) You should use the fact that $F(x,y)=0$ for all $x,y$ [currently you are ignoring that fact], (ii) It is better to write "$\frac{\partial f}{\partial u}$" rather than "$\frac{\partial f}{\partial x}$" to avoid confusion and to be consistent with the notation of the question [in particular, your $\frac{\partial x}{\partial x}$ is not quite correct], (iii) you forgot that differentiating $zx^{-\lambda}$ with respect to $x$ must use the product rule and that is where you get derivatives of $z$ in the picture. Can you solve the problem now? $\endgroup$ – Michael Jul 3 '16 at 3:32
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    $\begingroup$ Also, you can get rid of the formula "$\frac{\partial z}{\partial x} = -\frac{\partial f/ \partial x}{\partial f/\partial z}$" (I'm not sure what $\partial f /\partial x$ is even intended to mean). $\endgroup$ – Michael Jul 3 '16 at 3:35
  • $\begingroup$ @Michael I fixed it, but does it equal to 0 (the last expression)? $\endgroup$ – mvfs314 Jul 3 '16 at 4:04
  • $\begingroup$ I do not understand your comment above. What did you fix? What equals 0? If you can solve your own question now, one method is to answer your own question below. $\endgroup$ – Michael Jul 3 '16 at 15:55
  • $\begingroup$ I don't know what to do. $\endgroup$ – mvfs314 Jul 3 '16 at 18:39

$$df=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial v}dv=0$$ 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 $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}dx+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}dx-\frac{\lambda z}{x^{\lambda+1}}dx=0$$ $$\bigg(\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}\bigg)dx=0$$ $$\frac{1}{x^\lambda}\frac{\partial{z}}{\partial x}+\frac{1}{x^\lambda}\frac{\partial{z}}{\partial y}\frac{y}{x}-\frac{\lambda z}{x^{\lambda+1}}=0$$ $$x\frac{\partial{z}}{\partial x}+y\frac{\partial{z}}{\partial y}=\lambda z$$

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