In Problem 1, section A4 of this exam paper, the solution given below seem to use the fact that $$y\propto x^{\alpha} \Longrightarrow \frac{\delta y}{y} = \alpha \frac{\delta x}{x}$$ where $x$ and $y$ are two variables.
Not only is the meaning of $\delta$ ambiguous, I cannot find such usage online. What does the $\delta$ symbol mean here and why is the above statement correct?
Generally, $\delta x$ is notation for "a small change in $x$", and represents a quantity that is small relative to the actual value of $x$.
The statement $y \propto x^\alpha \implies \frac{\delta y}{y} \approx \alpha \frac{\delta x}{x}$ (note the approximately equal sign) can be found using either a Taylor series or a binomial expansion, for example:
$$\begin{eqnarray} y & \propto & x^\alpha \\ y & = & k x^\alpha \\ (y + \delta y) & = & k (x + \delta x)^\alpha \\ y \left( 1 + \frac{\delta y}{y} \right) & = & k x^\alpha \left(1 + \frac{\delta x}{x}\right)^\alpha \\ & = & k x^\alpha \sum_{n = 0}^\infty {\alpha \choose n}\left(\frac{\delta x}{x} \right)^n \\ & = & k x^\alpha \left(1 + \alpha \frac{\delta x}{x} + \frac{\alpha (\alpha - 1)}{2}\left(\frac{\delta x}{x} \right)^2 + \ldots \right) \\ & = & k x^\alpha\left(1 + \alpha \frac{\delta x}{x} + O\left(\left(\frac{\delta x}{x} \right)^2\right) \right) \\ k x^\alpha \left(1 + \frac{\delta y}{y}\right) & \approx & k x^\alpha \left(1 + \alpha \frac{\delta x}{x} \right) & \mbox{when } \frac{\delta x}{x} \mbox{ is small} \\ \implies \frac{\delta y}{y} & \approx & \alpha \frac{\delta x}{x} \end{eqnarray}$$
where $\alpha \choose n$ is the generalised binomial coefficient, and $O\left(\left(\frac{\delta x}{x}\right)^2\right)$ represents a collection of terms that are bounded by some constant multiple of $\left(\frac{\delta x}{x}\right)^2$ as $\delta x \rightarrow 0$, i.e. if $\delta x$ is "small enough" the error in the approximation is ignorable, and in fact it vanishes entirely in the limit.