I am trying to solve the following problem:
Transform the equation $(y-z)\frac{\partial z}{\partial x} + (y+z)\frac{\partial z}{\partial y} = 0$, taking $x=x(u,v)$ as a function and $u=y-z, v=y+z$ as independent variables.
The answer to the problem has to be: $\frac{\partial x}{\partial u}+\frac{\partial x}{\partial v}=\frac{u}{v}$
I have done other exercises involving change of varibles. However, all the exercises I have done had something in common: the function was $z$ (I mean, it was the function whose derivatives appeared in the given equation). Now that my function is $x$ I do not find a way to solve the problem.
It would be very helpful if someone could tell me the first steps of the correct approach to this kind of problem.
Thanks.
Idea: Use the chain rule to derive $z$ with respect to $u$ and $v$. Then, the expected terms $\frac{\partial x}{\partial u}$ and $\frac{\partial x}{\partial v}$ will appear. After you do that, use the equalities of the change of variables and the original equation to eliminate the derivatives of $z$.
Complete solution:
We have $$z=z(x,y)=z(x(u,v),y(u,v)).$$ Then, by the chain rule, $$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u},\qquad \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}\tag{1}$$
On the other hand, since $$u=y-z,\qquad v=y+z\tag{2}$$ we have $$\frac{\partial z}{\partial u}=\frac{\partial}{\partial u} \left(\frac{v-u}{2}\right)=-\frac{1}{2},\qquad \frac{\partial z}{\partial v}=\frac{\partial}{\partial v} \left(\frac{v-u}{2}\right)=\frac{1}{2}$$ $$\frac{\partial y}{\partial u}=\frac{\partial}{\partial u} \left(\frac{u+v}{2}\right)=\frac{1}{2},\qquad \frac{\partial y}{\partial v}=\frac{\partial}{\partial v} \left(\frac{u+v}{2}\right)=\frac{1}{2}$$ Substituting in $(1)$, we obtain $$-\frac{1}{2}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{1}{2}\frac{\partial z}{\partial y},\qquad \frac{1}{2}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{1}{2}\frac{\partial z}{\partial y}$$ and thus $$\frac{\partial z}{\partial y}=-\frac{\partial z}{\partial x}\cdot\left(\frac{\partial x}{\partial u}+\frac{\partial x}{\partial v}\right)\tag{3}$$ Now, substituting $(2)$ and $(3)$ in the original equation $$(y-z)\frac{\partial z}{\partial x} + (y+z)\frac{\partial z}{\partial y} = 0,$$ we get the desired result.