First, if $x=0$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $y\neq 0$. So we demand that $x= 0$ if and only if $y \neq 0$ (recall that $0/0$ is undefined). Next, if $x+y\neq 0$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $A$ gives

$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$

Then since the reciprocal of the reciprocal is our original function we have

$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $x\neq 0$ and $x \neq -y$.

As $x,y\in[0,1]$, we know that $x=-y$ if and only if $x=y=0.$ However, we have already excluded the case $x=0$ and $y=0$. Similarly, it follows that if $y=0$ then we demand that $x\neq 0$.

First, if $x=0$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $y\neq 0$. As $x,y\in[0,1]$, it follows that we exclude the case $x=y=0$. More specially, if $x=0$ then $y\neq 0$ and if $y=0$ then $x\neq 0$. Next, if $x+y\neq 0$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $A$ gives

$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$

Then since the reciprocal of the reciprocal is our original function we have

$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $x\neq 0$ and $x \neq -y$. Since $x,y\in[0,1]$, the second condition is only valid when $x=y=0$ which we excluded.

First, if $x=0$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $y\neq 0$. So we demand that $x= 0$ if and only if $y \neq 0$ (recall that $0/0$ is undefined). Next, if $x+y\neq 0$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $A$ gives

$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$

Then since the reciprocal of the reciprocal is our original function we have

$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $x\neq 0$ and $x \neq -y$.

First, if $x=0$ then $$\frac{x}{x+y}=\frac{0}{y}=0$$ provided $y\neq 0$. So we demand that $x= 0$ if and only if $y \neq 0$ (recall that $0/0$ is undefined). Next, if $x+y\neq 0$ then we may let $$A=\frac{x}{x+y}.$$ Taking the reciprocal of $A$ gives

$$\frac{1}{A}=\frac{x+y}{x}=1+\frac{y}{x}.$$

Then since the reciprocal of the reciprocal is our original function we have

$$A=\frac{1}{\left(\frac{1}{A}\right)}=\frac{1}{1+\frac{y}{x}}.$$ By our analysis above, we see that this is true provided $x\neq 0$ and $x \neq -y$.

As $x,y\in[0,1]$, we know that $x=-y$ if and only if $x=y=0.$ However, we have already excluded the case $x=0$ and $y=0$. Similarly, it follows that if $y=0$ then we demand that $x\neq 0$.