# How can I use Mathematica to solve the partial derivative of function of functions?

The implicit function $$z = z(x, y)$$ is determined by the equation $$f(\text{e}^z, 2z-x-y^2) = 0$$, and $$f$$ has a continuous partial derivative. How can I use Mathematica to find the partial derivatives of $$\partial_x z$$ and $$\partial_y z$$?

I am not sure this is what you want but..

tt = f[Exp[z[x, y]], 2 z[x, y] - x - y^2]


(* f(E^z(x,y),2 z(x,y)-x-y^2) *)

sol = Solve[{D[tt, x] == 0, D[tt, y] == 0}, {D[z[x, y], x],
D[z[x, y], y]}] // FullSimplify // First;

sol /. Derivative[a__][f][Exp[z[x, y]], 2 z[x, y] - x - y^2] :>
Derivative[a][f][X, Y] /. Exp[z[x, y]] -> Exp[z] // TableForm


Note that

D[z[x, y], x]/
D[z[x, y], y] /. sol


is equal to 1/(2y)

Another way that may be no more satisfying.

Your f[Exp[z[x, y]], 2 z[x, y] - x - y^2] == 0 can be rewritten as

eq = 2 z[x, y] - x - y^2 == g[Exp[z[x, y]]]


and Mathematica cannot solve this equation for z.

We can take derivatives.

D[eq[[1]], x] == D[eq[[2]], x]

Solve[%, Derivative[1, 0][z][x, y]]//Flatten

dzdx = Derivative[1, 0][z][x, y] /. %
(*-(1/(E^z[x, y] g'[E^z[x, y]] - 2))*)

D[eq[[1]], y] == D[eq[[2]], y]

dzdy = Derivative[0, 1][z][x, y] /. %
(*-((2 y)/(E^z[x, y] g'[E^z[x, y]] - 2))*)

dzdx/dzdy
(*1/(2 y)*)


If z weren't in both parts of f we could in many cases solve for it explicitly, but I don't know how to do it in this case.