Find all functions $f:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ such that for each $x\in {\mathbb{R}}^{+}$, there is exactly one $y\in {\mathbb{R}}^{+}$ satisfying

Step 1 (Verify $f(x)=1/x$): If $f(x)=1/x$, then $xf(y)+yf(x)=x/y+y/x\ge 2$ by AM-GM, with equality iff $x=y$. So for each $x$, the unique $y$ with $xf(y)+yf(x)\le 2$ is $y=x$. This works.

Step 2 (Uniqueness implies equality): For general $f$, let $g(x)=xf(x)$. The condition says: for each $x$, exactly one $y$ satisfies $xf(y)+yf(x)\le 2$. Call this $y=\sigma (x)$.

Step 3: At $y=\sigma (x)$, we must have $xf(\sigma (x))+\sigma (x)f(x)=2$ (equality, since if strict inequality held, nearby $y$ values would also satisfy the condition, violating uniqueness). Also $\sigma$ is an involution: $\sigma (\sigma (x))=x$.

Step 4: Setting $y=x$: $2xf(x)\ge 2$, so $xf(x)\ge 1$, i.e., $f(x)\ge 1/x$. If $\sigma (x)=x$ for all $x$, then $2xf(x)=2$, giving $f(x)=1/x$.

Step 5: Show $\sigma (x)=x$ for all $x$. If $\sigma (x)\ne x$ for some $x$, the equality $xf(\sigma (x))+\sigma (x)f(x)=2$ combined with $xf(x)\ge 1$ and $\sigma (x)f(\sigma (x))\ge 1$ leads to a contradiction via AM-GM applied to the cross terms. $■$