Derivatives of a function $y(x)$ which is implicitly defined by $F(x,y)=0$ I have a function $F(x,y)$ for which I have the analytic expression, and I also know all the partial derivatives $\frac{\partial^{(m+n)}F}{\partial x^m \partial y^n}$.
For each $x$, the equation $F(x,y) = 0$ has a unique solution for $y$, i.e. $F(x,y) = 0$ implicitly defines a function $y(x)$ which is continuous and differentiable.
I would like to express the derivatives $\frac{d^py}{dx^p}$ of the function $y(x)$ in terms of the known partial derivatives $\frac{\partial^{(m+n)}F}{\partial x^m \partial y^n}$.
How should I proceed?
 A: Since $F(x, y(x)) = 0$ is an identity between two functions, you may differentiate both sides $p$ times to obtain an identity
$$ \frac{\mathrm{d}^p}{\mathrm{d}x^p}F(x,y(x)) = 0. \tag{*} $$
It is not hard to see that

*

*The left-hand side of $\text{(*)}$ is a polynomial of $\frac{\partial^{i+j}F}{\partial x^i \partial y^j}(x, y)$, $i+j \leq p$, and $y', y'', \ldots, y^{(p)}$.


*Moreover, there is only one term in the left-hand side that involves $y^{(p)}$, namely $y^{(p)}(x)\frac{\partial F}{\partial y}(x, y(x))$.
So by rearranging, we may obtain a formula of the form
$$ y^{(p)}=-\frac{\text{[some expression involving $y,y',\ldots,y^{(p-1)}$]}}{\frac{\partial F}{\partial y}(x, y(x))}. $$
Now you may apply this relation recursively. For example, abbreviating $F^{(i,j)} = \frac{\partial^{i+j}F}{\partial x^i \partial y^j}(x, y(x))$ for brevity, then
\begin{align*}
y' &= - \frac{F^{(1,0)}}{F^{(0,1)}},
\\[1em]
y''
&= - \frac{y'^2 F^{(0,2)} + 2y'F^{(1,1)} + F^{(2,0)}}{F^{(0,1)}} \\
&= - \frac{F^{(2,0)} (F^{(0,1)})^2 - 2 F^{(1,0)} F^{(1,1)} F^{(0,1)} + F^{(0,2)} (F^{(1,0)})^2}{(F^{(0,1)})^3},
\\[1em]
y''' &= -\frac{3y'' F^{(1,1)} + (y')^3 F^{(0,3)} + 3(y'^2) F^{(1,2)} + 3 y' y'' F^{(0,2)} + 3 y'F^{(2,1)} + F^{(3,0)}}{F^{(0,1)}} \\
&= - \frac{\left( \begin{gathered}
F^{(3,0)} (F^{(0,1)})^{4}
-3 F^{(1,1)} F^{(2,0)} (F^{(0,1)})^{3}
-3 F^{(1,0)} F^{(2,1)} (F^{(0,1)})^{3} \\
+6 F^{(1,0)} (F^{(1,1)})^{2} (F^{(0,1)})^{2}
+3 (F^{(1,0)})^{2} F^{(1,2)} (F^{(0,1)})^{2} \\
+3 F^{(0,2)} F^{(1,0)} F^{(2,0)} (F^{(0,1)})^{2}
-F^{(0,3)} (F^{(1,0)})^{3} F^{(0,1)} \\
-9 F^{(0,2)} (F^{(1,0)})^{2} F^{(1,1)} F^{(0,1)}
+3 (F^{(0,2)})^{2} (F^{(1,0)})^{3}
\end{gathered}\right)}{(F^{(0,1)})^5}.
\end{align*}
As we see, the actual computation quickly goes out of hand as $p$ increases.
