If $f(x\cdot y)$ = $f(x). f(y)$ $\forall$ $x,y$ and $f(x)$ is continuous at $x = 1$. Prove following If $f(x\cdot y)$ = $f(x). f(y)$ $\forall$ $x,y$ and $f(x)$ is continuous at $x = 1$. Prove that $f(x)$ is continuous for all $x$ except at $x = 0$. Given $f(1)\ne0$.
$$f(1)=f(1)\cdot f(1)$$
$$f(1)(f(1)-1)=0$$
$$f(1)=0 \text { or } f(1)=1$$
As it is given $f(1)\ne0$, so $f(1)=1\tag{1}$
I know the condition for $f(x)$ to be continuous at all $x$ is:-
$$f(x^+)=f(x^-)=f(x)$$
Let's check the continuity at $x=0$
$$f(0)=f(0)f(0)$$
$$f(0)(f(0)-1)=0$$
Case $1$: $f(0)=1$
$$f(x\cdot0)=f(x)\cdot f(0)$$
$$f(0)(f(x)-1)=0$$
As $f(0)=1$, so 
$$f(x)-1=0$$
$$f(x)=1$$
One can clearly see that $f(x)$ is a continuous function $\forall x$. But in the question it is said that we have to prove $f(x)$ is continuous $\forall x$ except 0.
Case $2$: $f(0)=0$
$$f(x\cdot0)=f(x)\cdot f(0)$$
$$f(0)(f(x)-1)=0$$
We can't say that $f(x)=1$
How to proceed from here. I am totally stuck here and not finding  how to prove the given fact. Please help me in this.

Obviously $f(x+y)=f\left(x\cdot\left(1+\frac yx\right)\right)=f(x)\cdot f\left(1+\frac yx\right)$ where $x\not = 0$. 
So, we have $$\displaystyle\lim_{h\to 0} f(x\pm h)=f(x)\cdot f\left(1\pm\frac hx\right)$$
And since we are naturally observing the points where $x\not = 0$, $\frac hx\to 0$. And, $f(x)$ is continuous at $x\to 1$, $f\left(1\pm\frac hx \right) = f(1)=1$. So, using this, we have $$f(x\pm h)=f(x)\cdot 1$$ $$\implies f(x)\text{ is continuous}$$
