The given function is clearly continuous at all points except possibly at $x=\pm 1.$
For $f\left(x\right)$ to be continuous at $x=1,$ we must have
$\underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=f\left(1\right)$
$\Rightarrow \underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\alpha x^2+\beta =\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{x^2}$
$\Rightarrow \alpha +\beta =1$ … $\left(1\right)$
Now, for $f\left(x\right)$ to be differentiable at $x=1,$ we must have
$\underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$\Rightarrow \underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\alpha x^2+\beta -1}{x-1}=\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{1}{x^2}-1}{x-1}$ $\left(\because \alpha +\beta =1\therefore \beta -1=-\alpha \right)$
$\Rightarrow \underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\alpha x^2-\alpha }{x-1}=\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{1}{x^2}-1}{x-1}$
$\Rightarrow \underset{x\to 1^{-}}{\mathrm{l}\mathrm{i}\mathrm{m}}\alpha \left(x+1\right)=\underset{x\to 1^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{-2}{x^3}\Rightarrow 2\alpha =-2$
$\Rightarrow \alpha =-1$
Putting $\alpha =-1$ in $\left(1\right),$ we get $\beta =2$