Do you know the chain rule?
$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}\\ \frac{\partial u}{\partial y}=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y} $
or the inverse
$ \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\\ \frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t} $
You can use the second pair of rules, then solve for the desired quantities, or use the first pair of rules after obtaining $s(x,y), t(x,y)$ from the definitions of $x,y$.
Then you should proceed in a similar way for the second derivatives.
Edit
Going into the calculations:
\begin{align} \frac{\partial u}{\partial s}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\\ &=\frac{\partial u}{\partial x}e^s\cos t+\frac{\partial u}{\partial y}e^s\sin t=\\ &=x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\\ \frac{\partial u}{\partial t}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=\\ &=-\frac{\partial u}{\partial x}e^s\sin t+\frac{\partial u}{\partial y}e^s\cos t=\\ &=-y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y} \end{align}
This is true for every function $u$, so
\begin{align} \frac{\partial}{\partial s}&=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\\ \frac{\partial}{\partial t}&=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y} \end{align}
so we have
\begin{align} \frac{\partial^2u}{\partial s^2}&=\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)\\ \frac{\partial^2u}{\partial t^2}&=\left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\left(-y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y}\right) \end{align}
I suppose that from here you can go on without help.