Answer
The contour map of $f\left( {x,y} \right) = x$ consists of lines with equation $z=x=c$ for $c = \left( {..., - 2, - 1,0,1,2,...} \right)$ corresponding to the contour interval $1$.
Work Step by Step
If we write $z = f\left( {x,y} \right)$, the equation becomes $z=x$ or $x-z=0$. Note that this is an equation of the plane as discussed in Section 13.5. The plane intercepts the $x$-axis and the $z$-axis at $z=0$ and $x=0$, respectively. Since $y$-coordinate is arbitrary, the plane passes through the $y$-axis.
The level curves of $f\left( {x,y} \right) = x$ are the lines with equation $z=x=c$ for $c = \left( {..., - 2, - 1,0,1,2,...} \right)$ corresponding to the contour interval $1$.
