How does partial derivative work?

The multivariable chain rule states that, if $x = x(s,t)$, $y = y(s,t)$, and $u = u(x,y)$, then \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 t} &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} \end{align} To calculate the second derivatives \begin{align} \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial s}\right) &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s}\right) \\ &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial s}\right) + \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \frac{\partial y}{\partial s}\right)\\ &= \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x}\right) \frac{\partial x}{\partial s} + \frac{\partial u}{\partial x}\frac{\partial}{\partial s}\left(\frac{\partial x}{\partial s}\right) + \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y}\right) \frac{\partial y}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial}{\partial s}\left(\frac{\partial y}{\partial s}\right) \end{align} where the first step is the distribution of the derivative, the second is the product rule for differentiation.

Now, $$ \frac{\partial}{\partial s}\left(\frac{\partial x}{\partial s}\right) = \frac{\partial^2 x}{\partial s^2}, \qquad \frac{\partial}{\partial s}\left(\frac{\partial y}{\partial s}\right) = \frac{\partial^2 y}{\partial s^2} $$ and, using the multivaraible chain rule again \begin{align} \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x}\right) &= \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial s} = \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial \left(\tfrac{\partial u}{\partial x}\right)}{\partial y} \frac{\partial y}{\partial s} = \frac{\partial^2 u}{\partial x^2} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y \partial x} \frac{\partial y}{\partial s} \\ \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y}\right) &= \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial s} = \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial \left(\tfrac{\partial u}{\partial y}\right)}{\partial y} \frac{\partial y}{\partial s} = \frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y^2} \frac{\partial y}{\partial s} \end{align}

Supposing $\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}$ and substituting into $\frac{\partial^2 u}{\partial s^2}$, we have $$ \frac{\partial^2 u}{\partial s^2} = \frac{\partial^2 u}{\partial x^2} \left(\frac{\partial x}{\partial s}\right)^2 + 2\frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} \frac{\partial y}{\partial s} + \frac{\partial^2 u}{\partial y^2} \left(\frac{\partial y}{\partial s}\right)^2 + \frac{\partial u}{\partial x}\frac{\partial^2 x}{\partial s^2} + \frac{\partial u}{\partial y}\frac{\partial^2 y}{\partial s^2} $$

Why don't you calculate $\frac{\partial^2 u}{\partial s \partial t}$, $\frac{\partial^2 u}{\partial t^2}$ and see if you've improved your understanding?

The whole answer:

Let $v(s,t) = \frac{\partial u}{\partial x} e^s \cos t + \frac{\partial u}{\partial y} e^s \sin t$, where $x = x(s,t)$ and $y = y(s,t)$. Then \begin{align} \frac{\partial v}{\partial s} &= \frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x} e^s \cos t + \frac{\partial u}{\partial y} e^s \sin t\right)\\ &= \frac{\partial^2 u}{\partial s \partial x} e^s \cos t + \frac{\partial u}{\partial x} \frac{\partial}{\partial s}\big(e^s \cos t\big) + \frac{\partial^2 u}{\partial s \partial y} e^s \sin t + \frac{\partial u}{\partial x} \frac{\partial}{\partial s}\big(e^s \sin t\big)\\ &= \left(\frac{\partial^2 u}{\partial x^2} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y \partial x} \frac{\partial y}{\partial s}\right) e^s \cos t + \frac{\partial u}{\partial x}e^s \cos t \\ &\hskip2in + \left(\frac{\partial^2 u}{\partial x \partial y} \frac{\partial x}{\partial s} + \frac{\partial^2 u}{\partial y^2} \frac{\partial y}{\partial s}\right) e^s \sin t + \frac{\partial u}{\partial y}e^s \sin t \end{align} From the form of $v$ and your image, I'm assuming $x(s,t) = e^s \cos t$ and $y(s,t) = e^s \sin t$. In such case, \begin{align} \frac{\partial v}{\partial s} &= \frac{\partial^2 u}{\partial x^2} e^{2 s}\cos^2 t + 2 \frac{\partial^2 u}{\partial x \partial y} e^{2s} \cos t \sin t + \frac{\partial^2 u}{\partial y^2} e^{2 s}\sin^2 t + \frac{\partial u}{\partial x}e^s \cos t + \frac{\partial u}{\partial y}e^s \sin t \end{align} Finally, from the form of $v$, I think $v(s,t) = \frac{\partial u}{\partial s}$, but that information wasn't provided by the OP.

It is chain rule. Just write functions in a vector valued form nad use the chain rule.