Just add a normal vector of desired length.
The equation of the surface is $$g\colon (x,y)\mapsto \bigl(x,y,f(x,y)\bigr),$$ the normal vector is the cross product of the derivatives in respect to $x$ and $y$, namely $$n=\begin{pmatrix}1\\\0\\\f_x\end{pmatrix} \times\begin{pmatrix}0\\\1\\\f_y\end{pmatrix} =\begin{pmatrix}-f_x\\\\-f_y\\\1\end{pmatrix},$$ its length is $\sqrt{f_x^2+f_y^2+1}$. Now define for any real number $\epsilon$ the $\epsilon$-parallel surface $g_{\epsilon}$ of $g$ by $$g_{\epsilon}(x,y)=g(x,y)+\epsilon\cdot\frac{n(x,y)}{\|n(x,y)\|}$$
In our case $f(x,y)=x^2+y^2$. The normal vector here is $(-2x,-2y,1)$, its length is $\sqrt{4x^2+4y^2+1}$, hence $$g_{\epsilon}(x,y)=\begin{pmatrix}x\\\y\\\x^2+y^2\end{pmatrix} +\frac{\epsilon}{\sqrt{4x^2+4y^2+1}} \begin{pmatrix}-2x\\\\-2y\\\1\end{pmatrix}$$ will be the parallel surface in oriented distance $\epsilon$. ![epsilon=2]( First picture with $\epsilon=2$, second with $\epsilon=10$, where weird things are going on. ![](