Ygtjki

My question is similar to How to get the partial derivative of the inverse functions?

But they are different.

If we have a function $z=z(x,y)$, we can calculate the partial derivative $left.fracpartial^2zpartial x^2right|_y$. We can solve the original equation to obtain $x=x(z,y)$, and now we can also calculate the derivative $left.fracpartial^2xpartial z^2right|_y$.

I can directly calculate the relation between the two derivatives by hand. The result is
$$left.fracpartial^2zpartial x^2right|_y=-left(left.fracpartial xpartial zright|_yright)^-3cdotleft.fracpartial^2xpartial z^2right|_y.$$

What about higher-order derivatives? I think this is not a difficult job in MMA, but I cannot catch the point.

Here's another approach where I give Derivative a definition so that rules are not needed (it happens automatically). I'll use Michael's starting point:

eqn = x == f[z[x, y], y]

x == f[z[x, y], y]

and differentiate with respect to x:

deqn = D[eqn, x];
deqn //InputForm

1 == Derivative[1, 0][f][z[x, y], y]*Derivative[1, 0][z][x, y]

Solving for Derivative[1, 0][z][x, y] (which is $left. fracpartial zpartial x right|_y$ in your notation):

Derivative[1, 0][z][x, y] == 1 / Derivative[1, 0][f][z[x, y], y]

Let's turn this into a definition for Derivative:

Derivative[1, 0][z][x_, y_] = 1 / Derivative[1, 0][f][z[x, y], y];
Derivative[n_Integer?Positive, 0][z][x_, y_] := D[Derivative[1, 0][z][x, y], x, n-1]

Your first result can be obtained with:

Derivative[2, 0][z][x, y] //TeXForm

$-fracf^(2,0)(z(x,y),y)f^(1,0)(z(x,y),y)^3$

or:

D[z[x, y], x, 2] //TeXForm

$-fracf^(2,0)(z(x,y),y)f^(1,0)(z(x,y),y)^3$

Here's a table showing agreement with Michael's results:

Grid[
 Table[
 Derivative[n, 0][Inactive@z][x, y], Derivative[n, 0][z][x, y],
 n, 4
 ],
 Dividers -> All
] //TeXForm

$beginarrayc
hline
z^(1,0)(x,y) & frac1f^(1,0)(z(x,y),y) \
hline
z^(2,0)(x,y) & -fracf^(2,0)(z(x,y),y)f^(1,0)(z(x,y),y)^3 \
hline
z^(3,0)(x,y) & frac3
f^(2,0)(z(x,y),y)^2f^(1,0)(z(x,y),y)^5-fracf^(3,0)(z(x,y),y)f^(1,0)(z(x,y),y
)^4 \
hline
z^(4,0)(x,y) & -frac15 f^(2,0)(z(x,y),y)^3f^(1,0)(z(x,y),y)^7+frac10
f^(3,0)(z(x,y),y)
f^(2,0)(z(x,y),y)f^(1,0)(z(x,y),y)^6-fracf^(4,0)(z(x,y),y)f^(1,0)(z(x,y),y)^
5 \
hline
endarray$

This iterative method will give substitution rules up to the order equal to the maxorder. It's not a good idea to use x for both a variable and a function name, so I called it f. (For instance, if you want to replace the variable x by a number, Mathematica is also very likely to replace the x in the function x[z, y] by the number, which makes no sense. However the code below produces the right formula, if you use x[z[x, y], y] instead of f[z[x, y], y].)

iter[eq_, dz_, derivrules_] := #, #2, Join[derivrules, First@Solve[##]] &[
 D[eq, x] /. derivrules, D[dz, x]];
maxorder = 4;
drules = Last@Nest[iter, x == f[z[x, y], y], z[x, y], , maxorder];

Column[drules, Dividers -> All]

D[z[x, y], x, 3] /. drules