1. Rewrite the expression

Let $a = x^m, \qquad b = y^n.$ Since $x,y>0$ and $m,n$ are positive integers, we have $a,b>0$.

Then the given expression becomes $E=\frac{x^m y^n}{(1+x^{2m})(1+y^{2n})}=\frac{ab}{(1+a^2)(1+b^2)}.$

So the problem reduces to finding the maximum of $E(a,b)=\frac{ab}{(1+a^2)(1+b^2)}, \qquad a,b>0.$

2. Factor the expression

Observe that $E=\frac{a}{1+a^2}\cdot \frac{b}{1+b^2}.$

Thus we only need the maximum of the single-variable function $f(t)=\frac{t}{1+t^2}, \qquad t>0.$

3. Find the maximum of $f(t)$

Using the identity $1+t^2 \ge 2t$ for all $t>0$ (from AM-GM), we get $\frac{t}{1+t^2} \le \frac{t}{2t}=\frac12.$

Equality holds when $1+t^2=2t \iff (t-1)^2=0 \iff t=1.$

Hence, $\max_{t>0} \frac{t}{1+t^2} = \frac12.$

4. Apply this to both factors

Therefore, $E=\frac{a}{1+a^2}\cdot \frac{b}{1+b^2} \le \frac12\cdot\frac12=\frac14.$

Equality holds when $a=1, \quad b=1.$ That is, $x^m=1, \quad y^n=1.$ Since $x,y>0$, this gives $x=1, \quad y=1.$

5. Maximum value

So the maximum value is $\boxed{\frac14}.$

6. Check options

• A: $\frac12$ ❌
• B: $\frac14$ ✅
• C: $\frac{m+n}{6mn}$ ❌
• D: $1$ ❌

Thus the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.