So I am given the following: $$f(x,y)=x^y$$ $$u(x,y) = x + \ln y$$ $$v(x,y) = x - \ln y$$ and suppose that a new function defined as: $$g(u,v)=f(x(u,v),y(u,v))$$ and I am asked to find the partial derivative at a two points one of which is $\partial g(3,3)/\partial u$.
I am more interested in the way of thinking and approaching the question rather than the solution. Thanks a lot in advance :)
(1) Assuming $x(u,v)$ means the solution of $x$ satisfying $x+lny=u$ amd $x-lny=v$, we have $x(u,v)=\large{u+v\over2}$ and $y(u,v)=\large{e^{u-v\over2}}$.
Now $g=\large{({u+v\over2})^{e^{u-v\over2}}}$.
$\large{{\partial g\over\partial u}=({u+v\over2})^{e^{u-v\over2}}\cdot({\partial (e^{u-v\over2}ln({u+v\over2}))\over\partial u})}$
$\large{{\partial g\over\partial u}=({u+v\over2})^{e^{u-v\over2}}\cdot({{1\over2}e^{u-v\over2}ln({u+v\over2})}+{1\over u+v}e^{u-v\over2})}$.
(2) Assuming $x(u,v)$ means $\large{{\partial u \over\partial x}+{\partial v \over\partial x}}$
Then $x(u,v)=2$ and $y(u,v)=0$ and $g=2^0=1$ and $\large{{\partial g\over\partial u}}=0$