Solution:

The provided expressions are $x=a(\cos t+t \sin t)$and $y=a(\sin t-t \cos t)$

$x=a(\cos t+t \sin t)$

With respect to $t$ differentiating both the sides

$\frac{d x}{d t}=a\left(-\sin t+\frac{d}{d t} t \sin t\right)$

$\frac{d x}{d t}=a\left(-\sin t+t \frac{d}{d t} \sin t+\sin t \frac{d}{d t} t\right)$

$\frac{d x}{d t}=a(-\sin t+t \cos t+\sin t)$

$\Rightarrow \frac{d x}{d t}=a t \cos t$

And $y=a(\sin t-t \cos t)$

With respect to $t$ differentiating both the sides

$\frac{d y}{d t}=a\left(\cos t-\frac{d}{d t} t \cos t\right)$

$\frac{d y}{d t}=a\left(\cos t-\left(t \frac{d}{d t} \cos t+\cos t \frac{d}{d t} t\right)\right)$

$\frac{d y}{d t}=a(\cos t-(-t \sin t+\cos t))$

$\frac{d y}{d t}=a t \sin t$

So now $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a t \sin t}{a t \cos t}=\frac{\sin t}{\cos t}=\tan t$

Now again $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x} \tan t=\sec ^{2} t \frac{d}{d x} t$

$=\sec ^{2} t \frac{d t}{d x}=\sec ^{2} t \frac{1}{a t \cos t}$

$=\sec ^{2} t \frac{\sec t}{a t}=\frac{\sec ^{3} t}{a t}$