Given:  $f\left(x\right)=\left\{\begin{array}{ll}ax+b,& \text{ if }x\le 1\\ a{x}^2+c,& \text{ if }1<x\le 2\\ \frac{d{x}^2+1}{x},& \text{ if }x\ge 2\end{array}\right.$ is differentiable on $R$

So $f\left(x\right)$ should be continuous on $R$

Now continuity at $x=1$

$f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$

$\Rightarrow a+b=a+c\Rightarrow b=c ...\left(\mathrm{i}\right)$

Now continuity at $x=2$

$f\left(2^{-}\right)=f\left(2^{+}\right)=f\left(2\right)$

$\Rightarrow 4a+c=\frac{4d+1}{2}=8a+2c=4d+1 ...\left(\mathrm{ii}\right)$

Now 

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{ll}a,& \text{ if }x\le 1\\ 2ax,& \text{ if }1<x\le 2\\ \frac{d{x}^2-1}{x^2},& \text{ if }x\ge 2\end{array}\right.$

It is differentiable on $R$

So $f^{\text{'}}\left(1^{-}\right)=f^{\text{'}}\left(1^{+}\right)\Rightarrow a=2a\Rightarrow a=0 ...\left(\mathrm{iii}\right)$

$f^{\text{'}}\left(2^{-}\right)=f^{\text{'}}\left(2^{+}\right)\Rightarrow 4a=\frac{4d-1}{4}\Rightarrow d=\frac{1}{4} ...\left(\mathrm{iv}\right)$

Using Equations $\left(\mathrm{ii}\right), \left(\mathrm{iii}\right) \& \left(\mathrm{iv}\right)$

$c=b=1$

Now $ad-bc=0-1=-1$