Given that $a$ and $b$ are coprime integers, i.e $gcd(a,b)=1$ then for any multiplicative function $f$ will $f\left(\frac{a}{b}\right)$ be multiplicative? i.e Will following property hold
$$f\left(\frac{a}{b}\right) =f(a) \cdot f\left(\frac{1}{b}\right)$$
If it is true, please explain proof or intuition behind it.
If $f$ is only defined on the integers, then $f(a/b)$ is not defined for any integers $a,b$ except if $b$ divides $a$, i.e. $gcd(a,b)=b$. So your question does not really make sense.
What you can do if $f(a)\neq 0$ for all integers $a\neq 0$, is extend $f$ to all fractions by for all integers $f(a/b)=f(a)/f(b)$. This $f$ is still multiplicative on the rational numbers as for all fractions $\frac{a}{b},\frac{c}{d}\in\mathbb{Q}$ we have. (i.e. all the fractions.) $$f(\frac{a}{b}\cdot \frac{c}{d})=f(\frac{ac}{bd})=\frac{f(ac)}{f(bd)}=\frac{f(a)f(c)}{f(b)f(d)}$$ where the last equality is due to the multiplicity of $f$ on the integers.
I hope this helps.