Given that, $x=e^t\sin t,y=e^t\cos t \dots \left(1\right)$

At point $\left(1,1\right),1=e^t\sin t,1=e^t\cos t$

$\tan t=1 \Rightarrow t=\frac{\mathrm{\pi }}{4}$

On differentiating Equation $\left(1\right)$ w.r.t. $x$, we get

$\frac{dy}{dt}=e^t\left(\cos t-\sin t\right)$

and $\frac{dx}{dt}=e^t\left(\sin t+\cos t\right)$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos t-\sin t}{\cos t+\sin t}$

Again differentiating w.r.t. $x$, we get

$\frac{d^{2}y}{d{x}^2}=\frac{dx}{dt}\left(\frac{\cos t-\sin t}{\cos t-\sin t}\right)\frac{dt}{dx}$

$=\left[\frac{\left[\left(\cos t+\sin t\right)\left(-\sin t-\cos t\right)-\left(\cos t-\sin t\right)\left(-\sin t+\cos t\right)\right]}{{\left(\cos t+\sin t\right)}^2}\right]\frac{dt}{dx}$

$=\frac{-2}{{\left(\cos t+\sin t\right)}^2}.\frac{1}{e^t\left(\sin t+\cos t\right)}$

$= \frac{-2}{\left(e^t\cos t+e^t\sin t\right)}.\frac{1}{{\left(\cos t+\sin t\right)}^2}$

$=\frac{-2}{x+y}.\frac{1}{{\left(\cos t+\sin t\right)}^2 } \left[\mathrm{from}\mathrm{ }\mathrm{Equation}\left(1\right)\right]$

At $t=\frac{\pi }{4},x=1, y=1$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{-2}{1+1}.\frac{1}{{\left(\mathrm{c}\mathrm{o}\mathrm{s}\frac{\mathrm{\pi }}{4}+\mathrm{s}\mathrm{i}\mathrm{n}\frac{\mathrm{\pi }}{4}\right)}^2}$

$=\frac{-1}{\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]}=-\frac{1}{2}$