 Solution

Let $h=x^2+y^2+z-x$. Using Lagrange multipliers with $\nabla f=\lambda \nabla h$ givesThe third equation fixes $\lambda =1$. The first two then reduce to $2x-ay=1$ and $2y=ax$. If $a^2\ne 4$, their solution and the constraint giveAt $a=2$, the second equation requires $y=x$ and makes the first inconsistent; at $a=-2$, it requires $y=-x$ and gives the same inconsistency. Thus there is no constrained stationary point when $a=\pm 2$. The constraint gradient never vanishes because its vertical component is one, so the multiplier condition has not excluded any regular constrained stationary points.