One has $$ \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}\frac{d^n f(x)}{dx^n}\frac{d^n g(y)}{dy^n}=(exp(-(\frac{d^2}{dxdy}))[f(x)g(y)]-f(x)g(y) $$

the one-parameter corresponding group is $e^{t\frac{d^2}{dxdy}}=\Phi(t)$ acting on bivariate series as the sum you gave is

$$ \Phi(-1)[f(x)g(y)]-f(x)g(y) $$

 $\Phi(t)$ always converges with $t$ formal of $t$ real/complex on entire series. To integrate it, you can begin considering the functions $f(x)e^{y}\mapsto f(x+t)e^{y};\ e^xg(y)\mapsto e^{x}g(x+t)$. Hope this helps.

One has $$ \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}\frac{d^n f(x)}{dx^n}\frac{d^n g(y)}{dy^n}=(exp(-(\frac{d^2}{dxdy}))[f(x)g(y)]-f(x)g(y) $$

the one-parameter corresponding group is $e^{t\frac{d^2}{dxdy}}=\Phi(t)$ acting on bivariate series as the sum you gave is

$$ \Phi(-1)[f(x)g(y)]-f(x)g(y) $$  $\Phi(t)$ always converges with $t$ formal of $t$ real/complex on entire series. To integrate it, you can begin considering the functions $f(x)e^{y}\mapsto f(x+t)e^{y};\ e^xg(y)\mapsto e^{x}g(x+t)$. Hope this helps.

Note that, using eigenfunctions of the derivative to conjugate $\Phi(t)$, one has a differential expression of the "time-frequency" transform. Indeed $$ e^{-ay}\Phi(t)[f(x)e^{ay}]=f(x+at) $$ which acts on entire functions and corresponds to a dilation of the time by $a$ and a shift by $x$.