Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities $$ \tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2)). $$ its explicit form implies that $\mathrm{Pic}(\tilde{X}) \cong \mathbb{Z} \oplus \mathbb{Z}$. Furthermore, the exceptional divisor of the contraction $\tilde{X} \to X$ is the subvariety $$ E = \mathbb{P}_{\mathbb{P}^2}(\Omega_{\mathbb{P}^2}(1)) $$ and its embedding into $\tilde{X}$ is the relative double Veronese embedding. Finally, it is easy to check that the class of $E$ in $\mathrm{Pic}(\tilde{X})$ is primitive, therefore $$ \mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z}. $$

Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities $$ \tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2)). $$ its explicit form implies that $\mathrm{Pic}(\tilde{X}) \cong \mathbb{Z} \oplus \mathbb{Z}$. Furthermore, the exceptional divisor of the contraction $\tilde{X} \to X$ is the subvariety $$ E = \mathbb{P}_{\mathbb{P}^2}(\Omega_{\mathbb{P}^2}(1)) $$ and its embedding into $\tilde{X}$ is the relative double Veronese embedding. Finally, it is easy to check that the class of $E$ in $\mathrm{Pic}(\tilde{X})$ is equal to $$ 2H + 2h, $$ where $h$ is the hyperplane class of $\mathbb{P}^2$ and $H$ is the relative hyperplane class of $\mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2))$. Therefore $$ \mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}. $$