I know that
$$\begin{align*} \exp\left(\alpha\frac{d}{dx}\right)f(x)=f(x+\alpha)\,, \end{align*} \tag{1}$$
but I am looking for a definition for
$$\begin{align*} \exp\left(\alpha\frac{\partial}{\partial x}\frac{\partial}{\partial y}\right)f(x,y)\,, \end{align*} \tag{2}$$
so, the first question is: Is there a definition for the previous expression?.
Now, in the particular case where $f(x,y)$ is a product of two Gaussian functions centered in $y_{0}$ and $x_{0}$, that is $f(x,y)=\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right]$ (with $k_{1}$ and $k_{2}$ being real or complex constants), so, the second question is: Is it valid to apply equation (1) in order to displace first a Gaussian function?, I mean
$$\begin{align*} \exp\left(\alpha\frac{\partial}{\partial x}\frac{\partial}{\partial y}\right)\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right]\,, \end{align*} \tag{3}$$
taking $\hat{c} = \alpha\frac{\partial}{\partial x}$, the above expresión is $$\begin{align*} \exp\left(\hat{c}\frac{\partial}{\partial y}\right)\exp\left[-k_{2}(y-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \,, \end{align*} \tag{4}$$ Using equation (1) the $y$ variable is displaced $$\begin{align*} \exp\left[-k_{2}(\left[y+\hat{c}\right]-y_{0})^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \,, \end{align*} \tag{5}$$ or $$\begin{align*} =\exp\left[-k_{2}\left(\left[y+\alpha\frac{\partial}{\partial x}\right]-y_{0}\right)^{2}\right]\exp\left[-k_{1}(x-x_{0})^{2}\right] \ \end{align*} \tag{6}$$ then, the first exponential of above equation will become an operator that will act on the second exponential. I think that this procedure is wrong because if the exponentials of equation (3) are swapping, I would get
$$\begin{align*} \exp\left[-k_{1}(x-x_{0})^{2}\right] \exp\left[-k_{2}\left(\left[y+\alpha\frac{\partial}{\partial x}\right]-y_{0}\right)^{2}\right]\ \end{align*} \tag{7}$$
which clearly is not the same as the equation (6).