Taylor's polynomial for a 2 variables function

I must say the veracity of the affirmation : The second degree Taylor's polynomial of $\displaystyle f(x,y)=e^{xy}$ around $\displaystyle (0,1)$ is $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$.

My attempt : I didn't use the $\displaystyle u=xy$ substitution. Rather I did it the hard way : $\displaystyle p(x,y)=f(0,1)+\frac{\partial f(0,1)}{\partial x}(x)+\frac{\partial f(0,1)}{\partial y}(y-1)=1+0+0=1$.

Not sure I've done it right though. Could you second my answer?

( I just plotted the functions and $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$ seems to fit pretty well... meaning I made a mistake but I don't see where!)