\lim_{x\to 0^+}\frac{\sin x}{x}e^{-yx}=1=f(0,y) \Rightarrow f(x,y)=\displaystyle\left\{\begin{array}{cc}\displaystyle\frac{\sin x}{x}e^{-yx},~x>0 \cr~~~~~1, ~~~~~~x=0\end{array}