For the second problem,
$\frac { \partial ^ 2 u } { \partial x^2 }=\frac{\partial}{\partial x}(f'\frac{x}{r}) = \frac{x}{r}\frac{\partial}{\partial x}(f') + f' \frac{\partial}{\partial x} (\frac{x}{r})$
$ = \frac{x}{r}\frac{\partial}{\partial x}(f') + f' \frac{\partial}{\partial x} (\frac{x}{r})$
$ = \frac{x}{r} \frac{\partial (f')}{\partial r} \frac{\partial r}{\partial x} + f' (\frac{1}{r} - \frac{x}{r^2} \frac{\partial r}{\partial x})$
$ = \frac{x^2}{r^2} f'' + f' (\frac{1}{r} - \frac{x^2}{r^3})$
Edit: here is my earlier solution and I have added the missing terms.
$dz=f_1 du+f_2 dv$
Then $d^2z = f_{11} (du)^2 + f_{12} du \ dv + f_1 d^2u + f_{12} du \ dv + f_{22} (dv)^2 + f_2 d^2v$
You are missing $f_1 d^2u + f_2 d^2v$.
$f_1 d^2u = f_1 \ d (y dx + x dy) = f_1 (2 dx dy + y d^2x + x d^2y)$
Similarly, $f_2 d^2v = f_2 \ d (\frac{ydx - xdy}{y^2}) = f_2 \ [-\frac{2}{y^2} dx dy + \frac{1}{y} d^2x - \frac{x}{y^2} d^2y + \frac{2x}{y^3} (dy)^2]$
Or the way Graham Kemp explained,
$d^2 z = d^2x~\partial_x(z) + d^2y~\partial_y(z) + (d x)^2~\partial^2_x(z) + (d y)^2~\partial^2_y(z) + 2~dx~dy~\partial_y\partial_x(z)$
You can now differentiate each term and rearrange to get the same result. For example,
$d^2x~\partial_x(z) = d^2x~ (y f_1 + \frac{1}{y} f_2)$
I will take another term and let you work through the rest yourself.
$2 ~ dx ~ dy ~ \partial_y\partial_x(z) = 2 ~ dx ~ dy ~[\partial_y (y f_1 + \frac{1}{y}f_2]$
$ = 2 ~ dx ~ dy ~[f_1 - \frac{1}{y^2} f_2 + xy f_{11} - \frac{x}{y^3} f_{22}]$
Best Answer
The notation $\frac{\partial^2f}{\partial x\partial y}$ is the same as $$ \frac{\partial}{\partial x}\frac{\partial f}{\partial y}= \frac{\partial}{\partial x}\frac{2y}{x^2+y^2+3}\\ =2y\frac{\partial}{\partial x}\frac{1}{x^2+y^2+3}=2y\frac{-2x}{(x^2+y^2+3)^2}\\ =\frac{-4xy}{(x^2+y^2+3)^2}\\ $$ by the quotient rule and noting that $y$ is constant when we take the partial in $x$.