Find all functions such that $f(1+xf(y))=yf(x+y)$ where $x,y \in R^+$ 
Find all functions run over positive real numbers such that 
  $f(1+xf(y))=yf(x+y)$ where $x,y\in  R^+$

MY ANSWER:
Putting $x=y=0$,we get,
$f(1)=0$
Putting $x=0$ we get,
$f(1)=yf(y)$
or,$yf(y)=0$
or,$f(y)=0$ (since  $y\ne 0$., $y \in \mathbb R^+$)
Hence,$f(x)=0$ is the solution.
Is my answer and solution collect? If not then please tell me the proper answer and solution and where I have made the mistake!!
 A: For reference, here is CanVQ's solution taken from here. 

Replacing $x$ by $\frac{x}{f(y)}$ in $(1),$ we have
  $$f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)$$
  Next, we replace $y$ by $\frac{f(1+x)}{y}$ in $(2)$ to get
  $$y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\right)}+\frac{f(1+x)}{y}\right),\quad \forall x,\,y \in \mathbb R^+. \quad (3)$$
  From $(3),$ it follows that $f$ is surjective. Now, we will prove that $f$ is decreasing. Replacing $x$ by $x+z$ in $(1),$ we get
  $$f\big(1+(x+z)\cdot f(y)\big)=y\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (4)$$
  Replacing $y$ by $y+z$ in $(1),$ we also have
  $$f\big(1+x\cdot f(y+z)\big)=(y+z)\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad  (5)$$
  Dividing $(4)$ for $(5),$ side by side, we obtain
  $$\frac{f\big(1+(x+z)\cdot f(y)\big)}{f\big(1+x\cdot f(y+z)\big)}=\frac{y}{y+z},\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (6)$$
  Now, assume that there exists a pair $(y,\,z)$ such that $f(y+z)>f(y).$ In this case, by choosing $x=\frac{z\cdot f(y)}{f(y+z)-f(y)}$ vào $(6),$ we obtain $y=y+z,$ which is a contradiction. So we must have
  $$f(y+z) \le f(y),\quad \forall y,\,z \in \mathbb R^+. \quad (7)$$
  Now, we will prove that $f$ is injective. Assume that there are two numbers $a,\,b$ such that $f(a)=f(b).$ Replacing $y=a$ and $y=b$ in $(1)$ respectively in $(1),$ we get
  $$a\cdot f(x+a)=b\cdot f(x+b),\quad \forall x\in \mathbb R^+. \quad (8)$$
  From this, it follows that
  $$1+a(y-1)\cdot f(x+a)=1+b(y-1)\cdot f(x+b),\quad \forall x, \, y \in \mathbb R^+,\, y>1. \quad  (9)$$
  Plugging this into $f$ and using $(1),$ we get
  $$(x+a)\cdot f(x+ay)=(x+b)\cdot f(x+by),\quad \forall x,\,y \in \mathbb R^+ ,\, y>1. \quad (10)$$
  From $(10),$ it follows that, for any $x,\,y,\,z \in \mathbb R^+,\, y>1,$
$$1+(xz+az)\cdot f(x+ay)=1+(xz+bz)\cdot f(x+by). \quad (11)$$
  Again, we plug this into $f$ and using $(1).$ It follows that
  $$(x+ay)\cdot f(x+ay+az+xz)=(x+by)\cdot f(x+by+bz+xz)\quad  (12)$$
  for any $x,\,y,\,z \in \mathbb R^+$ and $ y>1.$ On the other hand, according to $(10),$ we also have
  $$\big[(x+xz)+a\big]\cdot f\big( (x+xz)+a(y+z)\big)=\big[(x+xz)+b\big] \cdot f\big((x+xz)+b(y+z)\big),$$
  or
  $$(x+xz+a)\cdot f(x+ay+az+xz)=(x+xz+b)\cdot f(x+ay+az+xz). \quad (13)$$
  Dividing $(12)$ for $(13),$ side by side, we obtain
  $$\frac{x+ay}{x+xz+a}=\frac{x+by}{x+xz+b},\quad \forall x,\,y,\,z \in \mathbb R^+,\, y>1. \quad (14)$$
  It is easy to deduce that $a=b$ here, so $f $ is injective. Now, replacing $x=y=1$ in $(1)$ with notice that $f $ is injective, we  have $f(1)=1.$ Since $f$ is strictly decreasing ($f$ is decreasing and injective), we have
  $$f(x)<1,\quad  \forall x>1.\quad (15)$$
  Now, let us consider the case $y>x.$ Replacing $y$ by $y-x$ in $(1)$ and using the above remark, we get
  $$f(y)=\frac{f\big(1+x\cdot f(y-x)\big)}{y-x}<\frac{1}{y-x},\quad \forall x,\,y \in \mathbb R^+,\, x<y. \quad  (16)$$
  In $(16),$ we let $x\to 0^+$ and obtain
  $$f(y) \le \frac{1}{y},\quad \forall y >0. \quad (17)$$ 
  Next, replacing $x$ by $x-1$ and $y$ by $f(y)$ in $(3),$ we get
  $$y=\frac{x-1}{f\left(\frac{f(x)}{f(y)}\right)}+\frac{f(x)}{f(y)},\quad \forall x ,\, y \in \mathbb R^+,\, x >1. \quad (18)$$
  Since $f\left(\frac{f(x)}{f(y)}\right) \le \frac{f(y)}{f(x)},$ from $(18),$ we deduce that
  $$y\ge \frac{(x-1)\cdot f(x)}{f(y)}+\frac{f(x)}{f(y)},$$
  or
  $$y\cdot f(y) \ge x \cdot f(x),\quad \forall x,\,y \in \mathbb R^+,\, x>1. \quad (19)$$
  Changin the position of $x$ and $y$ in $(19),$ we also have
  $$x\cdot f(x) \ge y\cdot f(y),\quad \forall x,\,y \in \mathbb R^+,\, y>1. \quad (20)$$
  From the inequalities $(19)$ and $(20),$ we can easily deduce that
  $$f(x)=\frac{k}{x},\quad \forall x>1. \quad (21)$$
  Now, taking $x=1$ in $(1)$ and using $(21),$ we have
  $$\frac{1}{1+f(y)}=\frac{y}{1+y},$$
  or
  $$f(y)=\frac{1}{y},\quad \forall y\in \mathbb R^+. \quad (22)$$
  Clearly, the function $f(x)=\frac{1}{x}$ satisfies our equation.

A: Here's a simple way to find the solutions.
Let us assume, that $f(x)$ solves the defining functional equation
$$f(1+xf(y))=y f(x+y)\tag{1}$$
Now putting $x=0$ turns out to be the only idea required here since the rest follows automatically.
Indeed, $x=0$ gives $f(1)=y f(y)$, or equivalently,
$$f(x)=f(1)/x\tag{2}$$
Inserting this into $(1)$ gives $$f(1+f(1)\frac{x}{y})\equiv \frac{f(1)}{1+f(1)\frac{x}{y}}=y f(1)\frac{1}{x+y} \tag{3}$$
Whence we have either $f(1)=0$ giving the trivial solution $f_1(x)=0$ for all $x$, or $f(1)\ne 0$ for which case we can divide by $f(1)$ and find
$$\frac{1}{1+f(1)\frac {x}{y}}=\frac{y}{x+y}=\frac{1}{1+\frac{x}{y}}\tag{4}$$ which requires $f(1)=1$ so that have narrowed down the candidate solution $(2)$ to the non-trivial solution $f_2(x)=\frac{1}{x}$.
Finally, it is very easy to show that $f_1(x)$ and $f_2(x)$ are indeed solutions.
And, by construction, there are no other solution.
I consider this to be a complete solution and would appreciate your criticism.
