1. By the chain rule,
[tex]\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}[/tex]
I'm going to switch up the notation to save space, so for example, [tex]z_x[/tex] is shorthand for [tex]\frac{\partial z}{\partial x}[/tex].
[tex]z_t=z_xx_t+z_yy_t[/tex]
We have
[tex]x=e^{-t}\implies x_t=-e^{-t}[/tex]
[tex]y=e^t\implies y_t=e^t[/tex]
[tex]z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}[/tex]
[tex]\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0[/tex]
Similarly,
[tex]w_t=w_xx_t+w_yy_t+w_zz_t[/tex]
where
[tex]x=\cosh^2t\implies x_t=2\cosh t\sinh t[/tex]
[tex]y=\sinh^2t\implies y_t=2\cosh t\sinh t[/tex]
[tex]z=t\implies z_t=1[/tex]
To capture all the partial derivatives of [tex]w[/tex], compute its gradient:
[tex]\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}[/tex]
[tex]\implies w_t=\dfrac1{\sqrt{-2t-t^2}}[/tex]
2. The problem is asking for [tex]\frac{\partial z}{\partial x}[/tex] and [tex]\frac{\partial z}{\partial y}[/tex]. But [tex]z[/tex] is already a function of [tex]x,y[/tex], so the chain rule isn't needed here. I suspect it's supposed to say "find [tex]\frac{\partial z}{\partial s}[/tex] and [tex]\frac{\partial z}{\partial t}[/tex]" instead.
If that's the case, then
[tex]z_s=z_xx_s+z_yy_s[/tex]
[tex]z_t=z_xx_t+z_yy_t[/tex]
as the hint suggests. We have
[tex]z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}[/tex]
[tex]x=s+t\implies x_s=x_t=1[/tex]
[tex]y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}[/tex]
Putting everything together, we get
[tex]z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)[/tex]
[tex]z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)[/tex]
