Am I going correctly? Let $f(x)$ be a polynomial in $x$ and let $a, b$ be two real numbers where $a \neq b$
Show that if $f(x)$ is divided by $(x-a)(x-b)$ then the remainder is
$\frac{(x-a) f(b)-(x-b) f(a)}{b-a}$
MY APPROACH:-
Let $Q(x)$ be quotient so that :
$(x-a)(x-b)Q(x)+{ Remainder }=f(x)$
L.H.S, $(x-a)(x-b) \cdot a(x)+ \cfrac{(x-a)f(b)-(x-b) f(b)}{b-a}$
So if we take $x=a$ : 
$ \text { Remainder }=\left.f(x)\right|_{x=a} $
$0+\frac{0-(x-b) f(b)}{(b-a)}=f(a) \\
f(b)=\frac{-(b-a) f(a)}{(x-b)}=\frac{(a-b){f}(a)}{x-b}$
again if we take $x=b,$ then $0+\frac{(x-a) f(b)}{b-a}=f(b)$
Substituting, $f(b)$ value in the equation then 
$\frac{(x-a) f(b)}{b-a}=\frac{(a-b) f(a)}{x-b}$

I think you have the right idea but have not set out your argument very clearly;  it looks as if you assume your conclusion about half way through.  A better lay out might be:
When dividing $f(x)$ by $(x-a)(x-b)$ you obtain,
$$ f(x) = (x-a)(x-b) q(x) + r(x) $$ where $r(x)$ is a polynomial of degree one or less. Thus $r(x) = \alpha x + \beta$ for values $\alpha$ and $\beta$.
Then substitute the value $x=a$ and $x=b$ in turn to derive,
$$ f(a) = r(a) = \alpha a + \beta, \quad f(b) = r(b) = \alpha b + \beta. $$
You an now solve for $\alpha$ and $\beta$ giving the result you wanted,
$$\alpha =\frac{f(b)-f(a)}{b-a} \text{ and } \beta = \frac{bf(a) - af(b)}{b-a}. $$
