If A is the position-vector representation of the point of contact of the plane on the surface; R, the generic position vector to the arbitrary point $\left(x\,y\,z\right)$; and N, a surface normal at the point of contact, then the vector equation for the tangent plane is $\left(\mathbf{R}-\mathbf{A}\right)·\mathbf{N}\=0$. This leads to

which can be rearranged to

$z\=f\left(a\,b\right)\+f_x\left(a\,b\right)\cdot \left(x-a\right)\+f_y\left(a\,b\right)\cdot \left(y-b\right)$ 

Guided by the Mathematical Solution, enter the appropriate equation, press the Enter key, and use the Context Panel's Solve option to isolate z.

An equation for a plane tangent to the surface $z\=f\left(x\,y\right)$ at the point $\left(a\,b\,f\left(a\,b\right)\right)$ is given in position-vector form by the TangentPlane command in the Student VectorCalculus package.

This form of the plane, essentially a parametric representation, can be interpreted as the equation

$z\=f\left(a\,b\right)\+f_x\left(a\,b\right)\cdot \left(x-a\right)\+f_y\left(a\,b\right)\cdot \left(y-b\right)$