If $y=\frac{1}{\sqrt{1+x^2}-x}$, then the value of $\left(1+x^2\right)^{\frac{3}{2}} \cdot \frac{d^2 y}{d x^2}$ is

1

$y=\frac{1}{\sqrt{1+x^2}-x}×\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}⇒\frac{\sqrt{1+x^2}+x}{1}$

$∴\frac{dy}{dx}=\frac{1}{2\sqrt{1+x^2}}×2x+1=\frac{y}{\sqrt{1+x^2}}$

$=\frac{d^2y}{dx^2}=\frac{\sqrt{1+x^2}\frac{dy}{dx}-y\frac{1}{2\sqrt{1+x^2}}.2x}{(1+x^2)}$

$⇒(1+x^2)^{\frac{3}{2}}=y(\sqrt{1+x^2}-x)=1$

Option 4 is correct.