Derivatives of $exp(f(x,y))$ I tried unsuccessfully to find this information in other posts, so I'll ask it here: 
The $n$-th derivative of $e^{f(x)}$ can be given in terms of the complete Bell polynomials:
$$
\begin{align}
\frac{d^n}{dx^n}e^{f(x)} &= e^{f(x)}B_n(f^{(1)}(x),...,f^{(n)}(x))\\
&= e^{f(x)}B_n^{(1)}(f)
\end{align}
$$
where $f^{(n)}(x) = \frac{d^n}{dx^n}f(x)$ and $B_n^{(1)}(f)$ indicates that we are considering the derivatives of $f$ with respect to its 1st argument.
I need the two-variable version of this formula, i.e. 
$$
\begin{align}
\frac{d^m}{dy^m}\frac{d^n}{dx^n}e^{f(x,y)} &= \frac{d^m}{dy^m}e^{f(x,y)}B_n(f^{(1,0)}(x,y),...,f^{(n,0)}(x,y))\\
&=\frac{d^m}{dy^m}e^{f(x,y)} B_n^{(1)}(f)\\
&=\sum_{k=0}^m \binom{m}{k}\frac{\partial^k e^{f(x,y)}}{\partial y^k}\frac{\partial^{m-k} B_n^{(1)}(f)}{\partial y^{m-k}}\\
&=e^{f(x,y)}\sum_{k=0}^m \binom{m}{k}B_k^{(2)}(f)\frac{\partial^{m-k} B_n^{(1)}(f)}{\partial y^{m-k}}
\end{align}
$$
So all I need to compute are the derivatives of $B_n(f^{(1,0)}(x,y),...,f^{(n,0)}(x,y))$ with respect to $y$.
We can consider the derivatives $f^{(j,0)}(x,y)$ as independent functions of $y$, so all in all I need a way to compute the terms:
$$
\frac{\partial^k}{\partial y^k}B_n(h_1(y),...,h_n(y))
$$
 A: 
We obtain
\begin{align*}
\color{blue}{\frac{d^m}{dy^m}}&\color{blue}{B_n\left(x_1(y),x_2(y),\ldots,x_n(y)\right)}\\
&=\frac{d^m}{dy^m}\sum_{k=1}^nB_{n,k}\left(x_1(y),x_2(y),\ldots,x_{n-k+1}(y)\right)\\
&=\frac{d^m}{dy^m}\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq  n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}}n!\prod_{l=1}^{n-k+1}
\frac{\left(x_l(y)\right)^{j_l}}{j_l!l!}\tag{1}\\
&=n!\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq  n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}}
\left(\prod_{l=1}^{n-k+1}\frac{1}{j_l!l!}\right)
\frac{d^m}{dy^m}\left(\prod_{l=1}^{n-k+1}\left(x_l(y)\right)^{j_l}\right)\\
&\,\,\color{blue}{=n!\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq  n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}}
\left(\prod_{l=1}^{n-k+1}\frac{1}{j_l!l!}\right)}\\
&\quad\qquad\color{blue}{\times\sum_{{j_t\geq 0,1\leq t\leq  n-k+1}\atop{\sum_{t=1}^{n-k+1}q_t=m}}
\binom{m}{q_1,q_2,\ldots,q_{n-k+1}}\prod_{t=1}^{n-k+1}\frac{d^{q_t}}{dy^{q_t}}(x_t(y))^{j_t}}\tag{2}
\end{align*}
which is admittedly not handy, but could be used for further analysis.

Comment:

*

*In (1) we use a representation  of  the  complete  exponential  Bell polynomial.


*In (2) we apply the general Leibniz rule.
A: Letting
\begin{align*}
\frac{\partial^{m+n}}{\partial x^m \partial y^n}e^{f(x,y)} = e^{f(x,y)}T_{m,n}\qquad\qquad  m\geq 0,n\geq 1\tag{1}
\end{align*}
we prove OPs recurrence relation
\begin{align*}
T_{m,n} &= \sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}T_{r,s}f^{(m-r,n-s)}(x,y)\qquad m\geq 0,n\geq1\tag{2}
\end{align*}

We obtain   from  (1) for  $m\geq 0, n\geq 1$:
\begin{align*}
\color{blue}{T_{m,n}}&=e^{-f(x,y)}\frac{\partial^{m+n}}{\partial  x^m\partial  y^n}e^{f(x,y)}\\
&=e^{-f(x,y)}\frac{\partial^{m+n-1}}{\partial  x^m\partial  y^{n-1}}\left(\frac{\partial}{\partial y}e^{f(x,y)}\right)\\
&=e^{-f(x,y)}\frac{\partial^{m}}{\partial  x^m}\frac{\partial^{n-1}}{\partial  y^{n-1}}\left(f^{(0,1)}(x,y)e^{f(x,y)}\right)\\
&=e^{-f(x,y)}\frac{\partial^{m}}{\partial  x^m}\sum_{s=0}^{n-1}\binom{n-1}{s}f^{(0,n-s)}(x,y)\frac{\partial s}{\partial  y^s}e^{f(x,y)}\tag{3}\\
&=e^{-f(x,y)}\sum_{s=0}^{n-1}\binom{n-1}{s}\frac{\partial^{m}}{\partial  x^m}\left(f^{(0,n-s)}(x,y)\frac{\partial ^s}{\partial  y^s}e^{f(x,y)}\right)\\
&=e^{-f(x,y)}\sum_{s=0}^{n-1}\binom{n-1}{s}\sum_{r=0}^m\binom{m}{r}f^{(m-r,n-s)}(x,y)\frac{\partial ^r}{\partial  x^r}\left(\frac{\partial ^s}{\partial  y^s}e^{f(x,y)}\right)\tag{4}\\
&=\sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}f^{(m-r,n-s)}(x,y)\left(e^{-f(x,y)}\frac{\partial ^{r+s}}{\partial  x^r\partial  y^s}e^{f(x,y)}\right)\\
&\,\,\color{blue}{=\sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}f^{(m-r,n-s)}(x,y)T_{r,s}}
\end{align*}
and the claim (2)  follows.

Comment:


*

*In (3) and (4) we apply the general Leibniz rule
A: The answer by Markus is definitely appreciated, but I may have found a simpler way.
We expect the solution to be in the form 
$$
\frac{\partial^{m+n}}{\partial x^m \partial y^n}e^{f(x,y)} = e^{f(x,y)}T_{mn},
$$
where $T_{mn} = \sum_{p,q}C^{mn}_{pq}f^{(p,q)}(x,y)$ is a polynomial in the partial derivatives of $f(x,y)$.
I have worked out a recursive definition of $T_{mn}$:
$$
\begin{align}
T_{m0} &= B_m(f^{(1,0)}(x,y),\dots,f^{(m,0)}(x,y))\\
T_{mn} &= \sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}T_{rs}f^{(m-r,n-s)}(x,y)\qquad n\geq1
\end{align}
$$
The only issue is that I don't have an actual proof, I found this result by generating a lot of examples with Mathematica and by staring at them long enough. 
