I prefer to maximize $z$, the visualization is more familiar, since $z$ is "up."
Think about how the plane $x+y+z=1$ meets the sphere $x^2+y^2+z^2=3$. Note that the plane has $x$-$y$ symmetry.
The highest point on the circle of intersection has $x=y$. So at the maximum we have $z^2+2x^2=3$, $z+2x=1$. Solve for $z$. We get $z=5/3$.
Added: The geometry is clear, but here is some algebraic backup. We have $x^2+y^2=3-z^2$ and $x+y=1-z$. Thus
Since $5+2z-3z^2=(5-3z)(1+z)$, and $(x-y)^2 \ge 0$, we conclude that
We are interested in positive $z$, so $5-3z \ge 0$, meaning that $z\le 5/3$, with equality when $x=y$.
Comment: About how to solve "this kind of problem," it depends how widely one interprets "this kind." The only fairly general kinds of techniques are numerical. For certain restricted classes of problems, Lagrange Multipliers can be useful. Here we took advantage of the symmetry. Special, perhaps, but many physically important problems have natural symmetries.