Answer
$$\frac{dw}{dt}=-\frac{\sin t\cos t-e^{2t}}{\sqrt{\cos^2t+e^{2t}}}$$
For $t=0$:
$$\frac{dw}{dt}=\frac{1}{\sqrt2}$$
Work Step by Step
Using the Chain rule we have:
$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}(\sqrt{x^2+y^2})\frac{d}{dt}(\cos t)+\frac{\partial}{\partial y}(\sqrt{x^2+y^2})\frac{d}{dt}(e^t)=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\partial x}(x^2+y^2)\cdot(-\sin t)+\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\partial y}(x^2+y^2)\cdot e^t=-\sin t\frac{2x}{2\sqrt{x^2+y^2}}+e^t\frac{2y}{2\sqrt{x^2+y^2}}=-\sin t\frac{x}{\sqrt{x^2+y^2}}+e^t\frac{y}{\sqrt{x^2+y^2}}$$
Expressing this in terms of $t$ we have:
$$\frac{dw}{dt}=-\sin t\frac{\cos t}{\sqrt{\cos^2t+e^{2t}}}+e^t\frac{e^t}{\sqrt{\cos^2t+e^{2t}}}=-\frac{\sin t\cos t-e^{2t}}{\sqrt{\cos^2t+e^{2t}}}$$
When $t=0$ we have:
$$\frac{dw}{dt}=\left.-\frac{\sin t\cos t-e^{2t}}{\sqrt{\cos^2t+e^{2t}}}\right|_{t=0}=-\frac{\sin0\cos0-e^{2\cdot0}}{\sqrt{\cos^20+e^{2\cdot0}}}=-\frac{0-1}{\sqrt{1+1}}=\frac{1}{\sqrt2}$$
