jmc

We can write $$\frac{xy}{x^2+y^2}=\frac{1}{\frac{x^2+y^2}{xy}}=\frac{1}{\frac{x}{y}+\frac{y}{x}}.$$Let $t=\frac{x}{y},$ so $\frac{x}{y}+\frac{y}{x}=t+\frac{1}{t}.$ We want to maximize this denominator.

Let $$f(t)=t+\frac{1}{t}.$$Suppose $0<t<u.$ Then $$\begin{array}{rl}f(u)-f(t)& =u+\frac{1}{u}-t-\frac{1}{t}\\ & =u-t+\frac{1}{u}-\frac{1}{t}\\ & =u-t+\frac{t-u}{tu}\\ & =(u-t)\left(1-\frac{1}{tu}\right)\\ & =\frac{(u-t)(tu-1)}{tu}.\end{array}$$This means if $1\le t<u,$ then $$f(u)-f(t)=\frac{(u-t)(tu-1)}{tu}>0,$$so $f(u)>f(t).$ Hence, $f(t)$ is increasing on the interval $[1,\infty ).$

On the other hand, if $0\le t<u\le 1,$ then $$f(u)-f(t)=\frac{(u-t)(tu-1)}{tu}<0,$$so $f(u)<f(t).$ Hence, $f(t)$ is decreasing on the interval $(0,1].$

So, to maximize $t+\frac{1}{t}=\frac{x}{y}+\frac{y}{x},$ we should look at the extreme values of $\frac{x}{y},$ namely its minimum and maximum.

The minimum occurs at $x=\frac{2}{5}$ and $y=\frac{3}{8}.$ For these values, $$\frac{xy}{x^2+y^2}=\frac{240}{481}.$$The maximum occurs at $x=\frac{1}{2}$ and $y=\frac{1}{3}.$ For these values, $$\frac{xy}{x^2+y^2}=\frac{6}{13}.$$Thus, the minimum value is $\boxed{\frac{6}{13}}.$