Xjyuk

Transform the following equation to new independent variables $u = x$, $v = x^2+y^2$

$$ yfracpartial zpartial x - x fracpartial zpartial y = 0 $$

Notce you don't have to actually solve the pde, the problem is to transform it. The official answer is $ fracpartial zpartial u = 0 $

My work:

$$ yfracpartial zpartial x = xfracpartial zpartial y longrightarrow frac1xfracpartial zpartial x = frac1yfracpartial zpartial y $$

$$ fracpartial zpartial y = fracpartial zpartial vfracpartial vpartial y longrightarrow fracpartial zpartial y = fracpartial zpartial v2y $$

Substituting, you get

$$ frac1x fracpartial zpartial x = 2fracpartial zpartial v $$

By the same procedure you can get

$$ fracpartial zpartial x = fracpartial zpartial u $$

Finally, I got

$$ frac1ufracpartial zpartial u = 2fracpartial zpartial v $$

Which is obviously not the official answer. Am I doing something wrong and if not, how do I finish the problem?

Thanks.

In your procedure the step $fracpartial zpartial x = fracpartial zpartial u$
does not true.

By chain rule:

$$fracpartial zpartial x = fracpartial zpartial ufracpartial upartial x+fracpartial zpartial vfracpartial vpartial x$$

$$fracpartial zpartial y = fracpartial zpartial ufracpartial upartial y+fracpartial zpartial vfracpartial vpartial y$$

$implies$ $$fracpartial zpartial x =fracpartial zpartial u(1)+fracpartial zpartial v(2x)=fracpartial zpartial u+2xfracpartial zpartial v$$ and

$$ fracpartial zpartial y =fracpartial zpartial u(0)+fracpartial zpartial v(2y)=2yfracpartial zpartial v$$

So $yfracpartial zpartial x - x fracpartial zpartial y = 0implies yfracpartial zpartial u+2yxfracpartial zpartial v-2xyfracpartial zpartial v=0implies fracpartial zpartial u=0$

$fracpartial zpartial x = fracpartial zpartial ufracd ud x+fracpartial zpartial vfracpartial vpartial x$

$$fracpartial zpartial x = fracpartial zpartial ucdot1 + fracpartial zpartial vcdot2x $$

Similarly,

$$fracpartial zpartial y = 0 + fracpartial zpartial vcdot2y$$

Now, $yfracpartial zpartial x - xfracpartial zpartial y = 0$

$yfracpartial zpartial u + 2xycdotfracpartial zpartial v -2xycdot fracpartial zpartial v=0implies yfracpartial zpartial u=0$

As, $y$ need not be necessarily $0$,

$$fracpartial zpartial u=0$$

$$partial zoverpartial x=duover dxpartial zoverpartial u+dvover dxpartial zoverpartial v=(1)partial zoverpartial u+(2x)partial zoverpartial v$$
$$partial zoverpartial y=duover dypartial zoverpartial u+dvover dypartial zoverpartial v=(0)partial zoverpartial u+(2y)partial zoverpartial v$$
$$0=ypartial zoverpartial x-xpartial zoverpartial y=yleft(partial zoverpartial u+2xpartial zoverpartial vright)-xleft(2ypartial zoverpartial vright)=ypartial zoverpartial u$$
Hence $partial zoverpartial u=0$

The more important point is that I don't think you understand how to use the chain rule in multivariable calculus - perhaps you should check how it works