Find an equation of the tangent plane to the given surface at the specified point.
z = ex2 - y2, (1, -1, 1)
First get the partial derivatives $\displaystyle \frac{\partial z}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}$.
Now construct the vector $\displaystyle \frac{\partial z}{\partial x} \, \vec{i} + \frac{\partial z}{\partial y} \, \vec{j} - \vec{k}$.
Substitute $\displaystyle x = 1$ and $\displaystyle y = -1$ into this vector. This gives a normal of the form $\displaystyle a \vec{i} + b \vec{j} - \vec{k}$ to the surface at $\displaystyle (1, -1, 1)$.
The equation of the tangent plane is therefore $\displaystyle ax + by - z = d$.
To get the value of $\displaystyle d$, substitute $\displaystyle x = 1$, $\displaystyle y = -1$ and $\displaystyle z = 1$.