jmc

We claim that the maximum value is 2. Note that for $x=y,$ $$\frac{(x+y{)}^2}{x^2+y^2}=\frac{4x^2}{2x^2}=2.$$The inequality $\frac{(x+y{)}^2}{x^2+y^2}\le 2$ is equivalent to $$(x+y{)}^2\le 2x^2+2y^2,$$which in turn simplifies to $x^2-2xy+y^2\ge 0.$ We can write this as $(x-y{)}^2\ge 0.$ This inequality holds, and since all our steps are reversible, the inequality $\frac{(x+y{)}^2}{x^2+y^2}\le 2$ also holds. Hence, the maximum value is $\boxed{2}.$