By the chain rule,
[tex]\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}[/tex]
We have
[tex]w=7ye^x-\ln z\implies\begin{cases}\dfrac{\partial w}{\partial x}=7ye^x\\\\\dfrac{\partial w}{\partial y}=7e^x\\\\\dfrac{\partial w}{\partial z}=-\dfrac1z\end{cases}[/tex]
and
[tex]\begin{cases}x=\ln(t^2+1)\\y=\tan^{-1}t\\z=e^t\end{cases}\implies\begin{cases}\dfrac{\mathrm dx}{\mathrm dt}=\dfrac{2t}{t^2+1}\\\\\dfrac{\mathrm dy}{\mathrm dt}=\dfrac1{t^2+1}\\\\\dfrac{\mathrm dz}{\mathrm dt}=e^t\end{cases}[/tex]
Putting everything together, we get
[tex]\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{14ye^xt}{t^2+1}+\dfrac{7e^x}{t^2+1}-\dfrac{e^t}z[/tex]
[tex]x=\ln(t^2+1)[/tex], so [tex]e^x=e^{\ln(t^2+1)}=t^2+1[/tex], and [tex]z=e^t[/tex], so [tex]\frac{e^t}z=1[/tex].
[tex]\dfrac{\mathrm dw}{\mathrm dt}=14yt+7-1[/tex]
[tex]\dfrac{\mathrm dw}{\mathrm dt}=14t\tan^{-1}t+6[/tex]
Then when [tex]t=1[/tex], the derivative has a value of [tex]\frac{7\pi}2+6[/tex].
