I'm trying to solve: $$\iint \ln(x+y^2)y^2 \, dx \, dy .$$ I substitute $x+y^2 = u$ and $y^2=v$. I calculate $\frac{dA_{xy}}{dA_{uv}}$ to be $$(1)\cdot(2y) - (2y)\cdot0 = 2y$$ ... which is equivalent to $2\sqrt{v}$ . I now need to solve: $$\iint \frac{\ln(u)\cdot v}{2 \sqrt{v}} \, du \, dv.$$ Setting both constants to $0$, this gives: $$\frac{v^{1.5}}{3}\cdot u\cdot(\ln(u)-1) $$ Substituting back $x+y^2 = u$ and $y^2=v$ we finally get: $$F(x,y) = \frac{y^3}{3}\cdot (x+y^2)\cdot(\ln(x+y^2)-1) $$ But this is not correct, at least when I check with a computer. Differentiating $F$ with $x$ and $y$ simply does not yield the original expression. CAS freeware program Geogebra gives us this behemoth: $$y^2\cdot\left(ln(x^{y^2}) - 1 \right)+\frac{2}{3}\cdot y^4\cdot\ln(x) + \frac{2y^6}{3x} + \frac{5}{3}y^4\cdot\frac{y^2+x}{x}. $$ This expression can surely be simplified, but when graphing it together with the original $\ln(x+y^2)\cdot y^2$ one sees that they are not identical.
What am I doing wrong in this derivation?
It has happened because the 2D indefinite integral is nonsense and this notation should be avoided imho. You probably mean something specific by this integral, but in that case, you should ask yourself what it is you want to find.
Let me try to guess. You are interested in a function $f$: $$ \frac{\partial^2}{\partial x\partial y}f(x,y) = y^2\ln(x+y^2). $$
You attempt to do a substitution $u=x+y^2$, $v=y^2$ and find function $g(u(x,y),v(y)) = f(x,y)$: $$ \frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}g(u,v)\right) = \frac{\partial}{\partial y}\left( \frac{\partial g(u,v)}{\partial u}\frac{\partial u}{\partial x} \right) = \frac{\partial}{\partial y}\left( \frac{\partial g(u,v)}{\partial u} \right) =\frac{\partial^2 g}{\partial u^2} \frac{\partial u}{\partial y} + \frac{\partial^2 g}{\partial u\,\partial v} \frac{\partial v}{\partial y} =\\ \left(2y\frac{\partial^2}{\partial u^2} + 2y\frac{\partial^2}{\partial u\,\partial v}\right)g = 2\sqrt{v}\left(\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial u\,\partial v}\right)g. $$
So we see that operator $\frac{\partial^2}{\partial x\partial y}$ doesn't transform into $\frac{\partial (u,v)}{\partial (x,y)}\frac{\partial^2}{\partial u\partial v}$ the same way as $dxdy$ transforms to $\frac{\partial (x,y)}{\partial (u,v)}dudv$. Whereas in 1D $\frac{\partial}{\partial q}$ is indeed $\frac{\partial p}{\partial q}\frac{\partial}{\partial p}$. This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.