Find an equation for the tangent plane for the function $f(x,y)=xe^{-xy}$, in the point $(1,0)$.
First I find the normal vector, which is given by $<1,0,e^{-xy}-ye^{-xy}> \times <0,1,-x^{2}e^{-xy}>$.
\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & e^{-xy}-ye^{-xy} \\ 0 & 1 & x^{2}e^{-xy} \end{vmatrix}
This gives me $\textbf{n}=<-e^{-xy}+yx^{-xy}, x^{2}e^{-xy},1>$
$\textbf{n} \bullet$(a vector which will force the equation to become zero in the specified point, because if the dot product of two vectors is zero they are orthogonal to each other. All the vectors that are orthogonal to the normal vector creates this plane)$=$equation of the tangent plane
$<-e^{-xy}+yx^{-xy}, x^{2}e^{-xy},1>\bullet<(x-1),(y-0),(z-1)>$ $=$ equation of the tangent plane
$<-e^{-xy}+yx^{-xy}, x^{2}e^{-xy},1>\bullet<(x-1),(y-0),(z-1)>$ $= z=1-x^{2}ye^{-xy}+xe^{-xy}+e^{-xy}+yx^{2}e^{-xy}+yx^{-xy}$
I don't understand what I am doing wrong.
The answer is: $z=x-y$ or $-x+y+z=0$
You need a specific normal vector. You're at the point $(x,y) = (1,0)$. Plug that into $\mathbf{n}$ and you should be fine. Also, your dot products should be set $=0$, not "$=$ to equation of tangent plane."