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 claim $$f(x+f(y+xy)) = (y+1)f(x+1)-1.$$
Notice that $P(x,\frac{y}{x+1})$ gives
$$
f(x+f(y))=(y+x+1)\frac{f(x+1)}{x+1}-1.
$$
To prove injectivity, assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$. Then comparing $P(x,\frac{a}{x+1})$ and $P(x,\frac{b}{x+1})$ we get
$$
(a+x+1)\frac{f(x+1)}{x+1}-1=(b+x+1)\frac{f(x+1)}{x+1}-1
$$
and hence $a=b$, i.e. $f$ is injective. Next $P(x,\frac{1}{f(x+1)})$ gives
$$
f(x+f(\frac{x+1}{f(x+1)}))=f(x+1)
$$
and by injectivity we get $f(\frac{x+1}{f(x+1)})=1$. So let $a$ be any such that $f(a)=1$, then $P(x,\frac{a}{x+1})$ gives
$$
f(x+1)=(a+x+1)\frac{f(x+1)}{x+1}-1
$$
and so solving for $f(x+1)$ we have
$$
f(x+1)=\frac{1}{a}(x+1).
$$
Plugging to the original equation we have
$$
f(x+f(y+xy)) = \frac{1}{a}(y+1)(x+1)-1.
$$
RHS is symmetric in $x,y$, so LHS must be too, hence we get a new claim $Q(x,y)$:
$$
x+f(y+xy)=y+f(xy+x).
$$
Now $Q(\frac{x}{2},1)$ gives
$$
\frac{x}{2}+f(\frac{x}{2}+1)=1+f(x).
$$
Since from previous $f(\frac{x}{2}+1)=\frac{1}{a}(\frac{x}{2}+1)$, we have found
$$
\frac{x}{2}+\frac{1}{a}\left(\frac{x}{2}+1\right)=1+f(x).
$$
Thus $f(x)$ is a linear function. Let $f(x)=cx+d$, then plugging into the original equation and comparing the coefficients we get $c=1,d=0$. So $f(x)=x$ is the only solution.
