Lagrange multipliers on a circle

Lagrange system.
$$\begin{cases} 3 = 2\lambda x \\ -1 = 2\lambda y \\ x^2 + y^2 = 5 \end{cases}$$

Resolution. From the first two equations, $x = \frac{3}{2\lambda}$ and $y = -\frac{1}{2\lambda}$, so $y = -\frac{x}{3}$. Substituting into the constraint:
$$x^2 + \frac{x^2}{9} = 5 \implies \frac{10x^2}{9} = 5 \implies x^2 = \frac{9}{2} \implies x = \pm \frac{3}{\sqrt{2}}.$$
The critical points are $\left(\frac{3}{\sqrt2}, -\frac{1}{\sqrt2}\right)$ and $\left(-\frac{3}{\sqrt2}, \frac{1}{\sqrt2}\right)$.

Evaluation.
$$f\left(\frac{3}{\sqrt2}, -\frac{1}{\sqrt2}\right) = \frac{9}{\sqrt2} + \frac{1}{\sqrt2} = \frac{10}{\sqrt2} = 5\sqrt2.$$
$$f\left(-\frac{3}{\sqrt2}, \frac{1}{\sqrt2}\right) = -\frac{9}{\sqrt2} - \frac{1}{\sqrt2} = -5\sqrt2.$$

Conclusion. Global maximum $5\sqrt2$ at the first point, global minimum $-5\sqrt2$ at the second. The constraint is a compact circle, so the extrema exist and are attained at these points.