Htydjtk

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 $left.frac{partial^2z}{partial x^2}right|_y$. We can solve the original equation to obtain $x=x(z,y)$, and now we can also calculate the derivative $left.frac{partial^2x}{partial z^2}right|_y$.

I can directly calculate the relation between the two derivatives by hand. The result is
$$left.frac{partial^2z}{partial x^2}right|_y=-left(left.frac{partial x}{partial z}right|_yright)^{-3}cdotleft.frac{partial^2x}{partial z^2}right|_y.$$

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

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

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 $left. frac{partial z}{partial x} right|_y$ 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

$-frac{f^{(2,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y)^3}$

or:

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

$-frac{f^{(2,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y)^3}$

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

$begin{array}{|c|c|}
hline
z^{(1,0)}(x,y) & frac{1}{f^{(1,0)}(z(x,y),y)} \
hline
z^{(2,0)}(x,y) & -frac{f^{(2,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y)^3} \
hline
z^{(3,0)}(x,y) & frac{3
f^{(2,0)}(z(x,y),y)^2}{f^{(1,0)}(z(x,y),y)^5}-frac{f^{(3,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y
)^4} \
hline
z^{(4,0)}(x,y) & -frac{15 f^{(2,0)}(z(x,y),y)^3}{f^{(1,0)}(z(x,y),y)^7}+frac{10
f^{(3,0)}(z(x,y),y)
f^{(2,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y)^6}-frac{f^{(4,0)}(z(x,y),y)}{f^{(1,0)}(z(x,y),y)^
5} \
hline
end{array}$