Let $f:(a,b)×ℝ→ℝ$ be a non zero and twice differentiable function. Let $c∈(a,b)$. Assume that $Δf=0$ (the Laplacian operator) and $f(c,y)=0$ for all $y∈ℝ$.
Assume that $$f(x,y)<0=f(c,y)$$ for all $(x,y)∈(a,c)×ℝ$ and $$f(x,y)>0=f(c,y)$$ for all $(x,y)∈(c,b)×ℝ$.
My question is: Can we consider the points $(c,y)$ as local minima or maxima for all $y∈ℝ$. If so, then the function $f$ must be a $constant=0$ according to the maximum principle for harmonic functions and this is a contradiction with the fact that $f$ is a non zero function.
A point $(x, y)$ is called a local minimum (maximum) if there is a open ball $B$ containing $(x, y)$ and $f(z, w) \geq (\leq) f(x, y)$ for all $(z, w)\in B$. So in your case $(c, y)$ is not a minimum (maximum) for all $y$.
Note that one candidate of $f$ is $f(x, y) = x-c$. It is harmonic and is not constant.