If $u = f(x,y)$, where $x=e^s \cos t$ and $y = e^s \sin t$, calculate $\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}$ 
If $ u = f(x,y)$, where $x=e^s \cos t$ and $y = e^s \sin t$, show that $$ \frac{\partial ^2u}{\partial x^2}  + \frac{\partial ^2u}{\partial y^2}= e^{-2s}\left[\frac{\partial ^2 u}{\partial s^2}+ \frac{\partial ^2 u}{\partial t^2}\right].$$

Currently, what I've done:
$$\begin{align}
f_x &= (f_s, f_t),
\\f_s &= e^s \cos t, 
\\f_t &= -e^s \sin t,
\\(f_x)_x &= ((f_s)_s, (f_t)_t),
\\(f_s)_s &= e^s \cos t,
\\(f_s)_t &= -e^s \sin t
\\(f_t)_s &= -e^s \sin t
\\ (f_t)_t &= -e^s \cos t
\\ \ 
\\f_y &=(f_s, f_t), 
\\f_s &= e^s \sin t, 
\\f_t &= e^s \cos t,
\\(f_y)_y &= ((f_s)_s, (f_t)_t),
\\(f_s)_s &= e^s \sin t,
\\(f_s)_t &= e^s \cos t
\\(f_t)_s &= e^s \cos t
\\ (f_t)_t &= -e^s \sin t
\end{align}$$
Could tutors over here advise me whether I am on the right track, as I have no idea how to proceed from here. Thanks :)
 A: You are mistaken at differentiating $f$, because \begin{gather} 
f_s =\frac{\partial  f}{\partial s}= \frac{\partial  f}{\partial x}\cdot\frac{\partial  x}{\partial s}+\frac{\partial  f}{\partial y}\cdot\frac{\partial  y}{\partial s}, \\ 
f_t =\frac{\partial  f}{\partial t}= \frac{\partial  f}{\partial x}\cdot\frac{\partial  x}{\partial t}+\frac{\partial  f}{\partial y}\cdot\frac{\partial  y}{\partial t}
\end{gather} 
Write $u=f(e^s \cos{t}, \, e^s \sin{t})$ and apply the chain rule calculating $\dfrac{\partial f }{\partial s},\, \dfrac{\partial f }{\partial s},\, \dfrac{\partial ^2 f}{\partial s^2}, \, \dfrac{\partial ^2 f}{\partial t^2}.$ For example, $$\frac{\partial f }{\partial s}=\frac{\partial  f}{\partial x}\cdot\frac{\partial x}{\partial  s}+\frac{\partial  f}{\partial y}\cdot\frac{\partial y}{\partial  s}=e^s \cos{t} \frac{\partial  u}{\partial x} +e^s\sin{t}\frac{\partial  u}{\partial y},\\
\frac{\partial f }{\partial t}=\frac{\partial  f}{\partial x}\cdot\frac{\partial x}{\partial  s}+\frac{\partial  f}{\partial y}\cdot\frac{\partial y}{\partial  s}=-e^s \sin{t} \frac{\partial  u}{\partial x} +e^s\cos{t}\frac{\partial  u}{\partial y}.$$
A: It helps when we get some grip on the things going on here. We are given a real-valued function
$$(x,y)\mapsto u(x,y)$$
defined in some part of the $(x,y)$-plane. Now the variables $x$ and $y$ are replaced by new variables $s$ and $t$ according to
$$x:=e^s\cos t\ ,\quad y=e^s\sin t\ .$$
(One may view this as a map of the $(s,t)$-plane into the $(x,y)$-plane.) In this way the function $u$ becomes a function of the new variables, called the pullback of $u$:
$$\tilde u(s,t):=u\bigl(x(s,t),y(s,t)\bigr)\ .$$
The partial derivatives of $\tilde u$ with respect to $s$ and $t$ are obtained by means of the chain rule:
$$\tilde u_s= u_x\,x_s+u_y\,y_s\ ,\quad \tilde u_t=u_x\,x_t+u_y\,y_t\ .$$
Now we are told to look at $\tilde D:=\tilde u_s^2 +\tilde u_t^2$. We get
$$\eqalign{\tilde D&=u_x^2(x_s^2+x_t^2)+2u_xu_y(x_s y_s+x_ty_t)+u_y^2(y_s^2+y_t^2)=\ldots\cr &=e^{2s}(u_x^2+u_y^2)\cr &=e^{2s}\, D\ .\cr}$$
