Let $\rm P(x,y): f \left(\frac{x}{y}\right) = f(x) + f(y) - f(x)f(y)$.$\rm P(x,1): f(1)=f(1)f(x)$. Case 1: $\rm f(1)\neq 0$.Then $\boxed {\rm f(x)=1,\forall x \in \mathbb{R_{+}}}$ Case 2:$\rm f(1)=0$.Then $\rm P(x,x): 0=2f(x)-f(x)f(x)\Leftrightarrow f(x)=0\,\,or \,\,f(x)=2$. If there exist $\rm k>0$ such that $\rm f(k)=2$ then : $\rm P(kx,k): f(x)=f(kx)+2-2f(kx)=2-f(kx)\Leftrightarrow f(x)=2-f(kx)$.So, $\rm 2-f(k\dfrac{x}{y})=2-f(kx)+2-f(ky)-f(x)f(y)\Leftrightarrow f(\dfrac{kx}{y})=f(kx)+f(ky)+f(x)f(y)\Rightarrow$ $\rm \overset{y=1}{\Rightarrow} f(kx)=f(kx)+f(ky)+0\Leftrightarrow f(k)=0$,contradiction.So $ {\rm f(x)=0 ,\forall x\in \mathbb{R_+}}$. So $\boxed {\rm f(x)=0 ,\forall x\in \mathbb{R_+}}$ or $\boxed {\rm f(x)=1,\forall x \in \mathbb{R_{+}}}$

Let $P(x,y)$ be the given assertion. Then $P(x,1)$ gives $f(1)(1-f(x))=0$. Hence $f\equiv 1$ is a solution. Assume there is a $c$ such that $f(c)\ne 1$. Then $f(1)=0$. This, together with $P(1,x)$, yields $f(1/x) = f(1)+f(x) - f(1)f(x) = f(x)$. Now, $P(x,x)$ yields $0=f(1) = 2f(x)-f(x)^2$ for all $x$. Finally, $P(y,1/y)$ yields $f(y^2) = f(y)+f(1/y)-f(y)f(1/y) = 2f(y)-f(y)^2 =0$. Hence $f(x)\equiv 0$ in this case.