Originally Posted by

**arbolis** I see. I get $\displaystyle \frac{\partial f^2(x,y)}{\partial x ^2}=y^2 e^{xy}$, so evaluated in $\displaystyle (0,1)$ and divided by $\displaystyle 2$ and multiplied by $\displaystyle x^2$ gives $\displaystyle \frac{x^2}{2}$.

However I get that $\displaystyle \frac{\partial f^2(x,y)}{\partial y ^2}=x^2 e^{xy}$, evaluated in $\displaystyle (0,1)$ it gives $\displaystyle 0$. Hence $\displaystyle p(x,y)=1+\frac{x^2}{2}$. ( I checked what you told me, that the mixed partial derivatives are 0, that is, $\displaystyle f_{xy}(0,1)=0$)

I feel like I'm missing something obvious. I'll go to sleep and come back tomorrow. Thanks for your help CaptainBlack.