Answer
Please see image below.
Work Step by Step
1. No x in the equation, no intercept with the x-axis.
2. Set $x=0$. The trace is a line ($z=1+y$). Find intercepts with the y and z-axes.
$f(0,0)=1$, so the z intercept is at $(0,0,1)$
When $z=f(x,y)=0$,
$0=1+y$
$y=-1$. so the y-intercept is at $(0,-1,0).$
3. Sketch a line through the intercepts (in the yz plane),
4. Sketch lines through the intercepts, parallel to x
5. The result is a plane.
