show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$ 
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that  $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.

I think it might be useful to prove that f is injective and then make a substitution. It should be possible to prove $f$ is surjective too. It might also help to prove additional properties of f such as multiplicativity, that f has infinitely many fixed points, that f satisfies the functional equation $f(x+y) = f(x)+f(y)$ for all positive real numbers x,y, etc. First substituting $x=y$ gives $f(x+f(x+x^2)) = (x+1)f(x+1)-1.$ Swapping $x$ and $y$ gives $f(y+f(x+xy)) = (x+1)f(y+1)-1.$ Substituting $x=1$ into the latter equation gives $f(1+f(2)) = 2 f(2) - 1.$ In general for any positive integer $n, f(n+f(n+1)) = 2f(n+1)-1$. Suppose $f(x)=f(y).$ Then  It might be useful to make the substitution $u=y+1, v= x+1$. Then we have $f(u-1 + f(u(v-1))) = uf(v) - 1$ for all $u,v > 1.$
If $f(\dfrac{x+y}2) = \dfrac{f(x)+f(y)}2$ for $x,y \in \mathbb{R}_{\ge 0}$, then letting $g(x) = f(x) - f(0)$ we get $g((x+y)/2) = (g(x)+g(y))/2$ for all $x\in \mathbb{R}^+$ so we may assume $f(0) = 0$. Then $f(x) = f(2x)/2$ for all nonnegative x and so $f(x+y) = (f(2x)+f(2y))/2 = f(x)+f(y)$, implying that $f(nx) = nf(x)$ for all rationals $n > 0$ and $x \ge 0$.

Let $P(x,y)$ be the assertion, as usual.
$P(1,x/2)\implies f(1+f(x))=(1+x/2)f(2)-1$. Thus if $f(a)=f(b)$, $a=b$ i.e. $f$ is injective.
Now we put the fairly obvious $P(x,\frac{1}{f(x+1)})$ to get $f(x+f\left(\frac{x+1}{f(x+1)}\right))=f(x+1)$ which, by injectivity, implies $f\left(\frac{x+1}{f(x+1)}\right)=1$. Again by the injectivity $\frac{x+1}{f(x+1)}$ is fixed for all $x$, so $f(x)=cx$ for some constant $c$. A quick plug back into the original problem yields $c=1$, as wanted.
