$13.$ Consider the change of variables
$$x = e^{−s} \sin t,\space y = e^{−s} \cos t, \space \text{such that} \space u(x,y) = v(s,t)$$
(i) Use the chain rule to express $∂v/∂s$ and $∂v/∂t$ in terms of $x$, $y$, $∂u/∂x$ and $∂u/∂y$.
(ii) Find, similarly, an expression for $∂^2v/∂t^2$.
(iii) Hence transform the equation $$y^2 \frac{∂^2u}{∂x^2} −2xy \frac{∂^2u}{∂x∂y} +x^2 \frac{∂^2u}{∂y^2}=0$$ into a partial differential equation for $v$.
Sorry for the poor formatting, the question is also question $13$ on http://www.damtp.cam.ac.uk/user/examples/A3d.pdf
I have done the first parts but it seems that I have gone wrong as I cannot see how they fit together to transform the final equation. All help is appreciated.
Your step 1 is correct. Let's recap it as $$v_s = -xu_x-yu_y,\quad v_t = yu_x-xu_y$$ To calculate $v_{tt}$, insert the formula for $v_t$ into itself: $$v_{tt} = y(yu_x-xu_y)_x-x(yu_x-xu_y)_y $$ This way you are not mixing $xy$ and $ts$ derivatives. The rest of computation is straightforward: $$v_{tt} = y^2u_{xx}-2xyu_{xy}+x^2 u_{yy} - yu_y - xu_x $$ The third step should be easy now.