Assume that we have differential equation $$f'(x)=g(f(x))$$ and we know that function $h(x,c)$ is the solution (with parameter) to it, then what can be said of the solution for the equation $$-f'(x)=g(f(x))?$$
It would seem that if the original equation the derivative is positive (negative), ie. $h(x,c)$ is increasing (decreasing), then in $-f'(x)=g(f(x))$ it is negative (positive) and thus the answer should be something like $-h(-x,c)$, but I am not certain. When is the answer $\frac{1}{h(x,c)}?$
Lets write $h_c(x) = h(x,c)$ to distinguish the parameter and the variable more clearly, then $h_c^\prime(x) = g(h_c(x))$. Furthermore define $k_c(x) = h_c(-x)$, then $$k_c^\prime(x) = \frac{\mathrm{d}}{\mathrm{d}x}h_c(-x) = -h_c^\prime(-x) = -g(h_c(-x)) = -g(k_c(x)).$$ Hence $k_c(x)=h_c(-x)$ solves the equation $-f^\prime(x)=g(f(x))$.
Again suppose $h_c^\prime(x) = g(h_c(x))$ and define $l_c(x) = h_c(x)^{-1}$, then $$l_c^\prime(x) = \frac{\mathrm{d}}{\mathrm{d}x}h_c(x)^{-1} = -h_c(x)^{-2}h^\prime_c(x) = -h_c(x)^{-2}g(h_c(x)) = -l_c(x)^2g(l_c(x)^{-1}).$$
Thus, $l_c(x) = h_c(x)^{-1}$ would be the solution to the equation $f^\prime(x) = \hat{g}(f(x))$, where $\hat{g}(y) = -y^2g(y^{-1})$.