How do I go from system $(1)$ to system $(2)$ below?
\begin{gather} \begin{aligned} S &= K e^{x}\\[7pt] V(S,t) &= K v(x, \tau)\\ \tau &= (T-t) \frac{\sigma^{2}}{2} \end{aligned}\tag{1} \end{gather}
\begin{gather} \begin{aligned} \frac{d}{dt}V &= -K \frac{\sigma^{2}}{2} \frac{d}{d \tau} v\\ \frac{d}{dS}V &= \frac{K}{S} \frac{d}{dx} v \end{aligned}\tag{2} \end{gather}
Let's start with the first one: $$V(S,t) = K v(x, \tau).\tag{1}$$ Differentiating each side, using the chain rule: $$ \frac{d}{dt}V(S,t) = \frac{d}{dt}Kv(x,\tau)= K\frac{dv}{d\tau}(x,\tau)\cdot\frac{d\tau}{dt}(t).\tag{2}\\ $$
Note that $x$ does not depend on $t$, which is why we only have to consider $\frac{dv}{d\tau}$ above. Now, using $$\tau= (T-t) \frac{\sigma^{2}}{2}$$ we can obtain $$\frac{d\tau}{dt} = -\frac{\sigma^{2}}{2},$$ and so $(2)$ becomes $$\boxed{\frac{d}{dt}V(S,t) = -K\frac{\sigma^{2}}{2}\frac{dv}{d\tau}(v,\tau).}$$
For the second, we differentiate $(1)$ with respect to $S$, again using the chain rule. This time we only have to consider $\frac{dv}{dS}$ because $\tau$ does not depend on $S$. $$\frac{d}{dS}V(S,t) = \frac{d}{dS}Kv(x,\tau) = K\frac{dv}{dx}(x,\tau)\cdot \frac{dx}{dS}(S).\tag{3}$$ Now, to compute $\frac{dx}{dS}$ we differentiate implicitly: \begin{align} S &= Ke^{x}\\ 1 &= Ke^{x}\frac{dx}{dS}\\ \end{align} Solving for $\frac{dx}{dS}$, making using of $S = Ke^{x}$, we get $$\frac{dx}{dS} = \frac{1}{Ke^{x}}=\frac{1}{S}.\tag{4}$$ Plugging $(4)$ into $(3)$ gives us $$\boxed{\frac{d}{dS}V(S,t) = \frac{K}{S}\frac{dv}{dx}(x,\tau).}$$