= Solution
For this <equal-strength power-law mutual repression>, the logarithmic gains and their product at equilibrium are
$$
\epsilon_f=-\frac{ny}{1+y},\qquad \epsilon_g=-\frac{mx}{1+x},\qquad
P=\epsilon_f\epsilon_g=\frac{mnxy}{(1+x)(1+y)}.
$$
The <mutual repression stability criterion> immediately excludes <multistability> if $mn\le1$, since $P<mn$ at every finite positive equilibrium. The common prefactor $\lambda$ gives a stronger exclusion: \b[if either exponent is at most one, multistability is impossible for every $\lambda$], including some cases with $mn>1$.
To prove the additional cases, suppose $m\le1<n$. At equilibrium the two expressions for $\lambda$ imply
$$
\frac{x}{(1+x)^m}=\frac{y}{(1+y)^n}.
$$
Put $a=x/(1+x)$ and $b=y/(1+y)$. Because $m\le1$, the left side is at least $a$, so $a\le b(1-b)^{n-1}$. It follows that
$$
P=mnab\le mn b^2(1-b)^{n-1}
\le\frac{4mn(n-1)^{n-1}}{(n+1)^{n+1}}<1.
$$
The maximum in the second inequality occurs at $b=2/(n+1)$. Its strict bound follows from $4n/(n+1)^2\le1$, $m\le1$, and $((n-1)/(n+1))^{n-1}<1$. Interchanging the components proves the case $n\le1<m$; the case with both exponents at most one was already covered by $mn\le1$. Every root of $F(x)=f(g(x))-x$ therefore has $F'<0$. Since $F$ begins positive and ends negative, there is exactly one root: two downward crossings would require another crossing with nonnegative derivative between them. The unique equilibrium is stable.
For completeness, neither exponent can be relaxed above one in this exclusion. If $m>1$ and $n>1$, take $\lambda$ sufficiently large. On $x\in[\lambda/2,\lambda]$, $g(x)=O(\lambda^{1-m})\to0$, so $f(g(x))=\lambda(1+o(1))$ maps that interval into itself. It has a fixed point with
$$
x\sim\lambda,\qquad y\sim\lambda^{1-m},\qquad P\to0.
$$
It is a <stable equilibrium>. By exchanging the components, there is a second stable fixed point with $y\sim\lambda$ and $x\sim\lambda^{1-n}$. These two regions are disjoint for large $\lambda$, proving <bistability> for some production strength. Hence the full answer is
$$
\boxed{\text{No multistability for any }\lambda>0\quad\Longleftrightarrow\quad m\le1\ \text{or}\ n\le1.}
$$
For example, $m=1/2,n=3$ has $mn>1$ but still a unique stable equilibrium for all $\lambda$. Thus the product-only necessary condition for strong feedback is not sufficient when both production prefactors are constrained to be equal.
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