Let $\mathbf{x} = \begin{bmatrix} x\\ y\\ z \\ \end{bmatrix}$ and $Q=\begin{bmatrix} a & d/2 & f/2\\ d/2 & b & e/2\\ f/2 & e/2 & c \\ \end{bmatrix}$.
Define a plane projective curve to be the zero locus of $q(\mathbf{x})=\mathbf{x}^tQ\mathbf{x}$.
Consider the following equivalence relation on the set of all such curves: two curves are equivalent iff there is a linear change of variables that maps one curve to the other one.
How to describe the set of equivalence classes under this relation? The only thing I can come up with is to write $\widetilde{\mathbf{x}}=T\mathbf{x}$ for some $3\times 3$ matrix $T$ so that $q(\widetilde{\mathbf{x}})=\mathbf{x}(T^tQT)\mathbf{x}$. Eventually I should get that there are $3$ equivalence classes, and they are the zero loci of $x^2,xy,xz-y^2$.
The transformation $Q \rightsquigarrow T^t Q T$ is just the transformation of $Q$ determined by $T \in M(3, \Bbb F)$, regarded as a (symmetric) bilinear form on $\Bbb F^3$, so we are effectively classifying those forms up to that action (and then discarding the bilinear forms that do not give curves).
The classification of symmetric forms depends on the choice of underlying field $\Bbb F$, but the given solution suggests that $\Bbb F = \Bbb C$, in which case the classification is given precisely by the rank of the matrix. More concretely, a complete set of representatives is $$\pmatrix{0& &\\&0&\\& &0}, \quad \pmatrix{1& &\\&0&\\& &0}, \quad \pmatrix{1& &\\&1&\\& &0}, \quad \pmatrix{1& &\\&1&\\& &1}.$$
The bilinear forms resulting from these normal forms may need to transformed again to produce the bilinear forms given in the classification: For example, if we take $Q$ to be the third matrix, the corresponding plane curve is $x^2 + y^2$, but an appropriate choice of linear transformation (linear change of coordinates) $T$ transforms that matrix to $$\tilde Q := T^t Q T = \pmatrix{&\frac{1}{2}&\\\frac{1}{2}& &\\& &0},$$ which corresponds to $xy$, the second plane curve in the given classfication.