You make your solution feel "fancier", you could start with a more coordinate free approach to the dual curve. For example, try to identify the line bundle on your curve which embeds it into the dual plane. E.g.: let $X \subset \mathbb{P}^2$ be your curve. Then define $X' \subset X \times \mathbb{P}^{2*}$ to be the set of pairs $(x, H)$ where the line $H$ is contained in the tangent space to $X$ at $x$. When $X$ is smooth at least, then this may be identified with the projectivization of the dual of the normal bundle $N_{X/\mathbb{P}^2}$.
Then think about the map $X' \rightarrow X \times \mathbb{P}^{2*} \rightarrow \mathbb{P}^{2*}$. This is the dual curve embedding. The morphism comes from the inclusion of $N^* \subset \Omega_{\mathbb{P}^2}|_X \subset \mathcal{O}(-1)^3$.
At some point, you have to do computations in local coordinates - but setting it up like this may help a little.
Example of local computation: Locally, the curve looks like $(x(t), y(t), z(t))$. The map to the dual plane is given by $(x(t), y(t), z(t)) \times (x'(t), y'(t), z'(t))$ (cross product! giving the normal to the tangent plane in $\mathbb{C}^3$). If the curve has a simple inflection point, then locally it looks like $(t, t^3, 1)$ and the map to the dual plane is given by $(3t^2, 1, 2t^3)$ - a curve with a simple cusp.
