If $$6(\sec^2 59^\circ - \cot^2 31^\circ) - \frac{2}{3}\sin 90^\circ - 3 \tan^2 56^\circ y \tan^2 34^\circ = \frac{y}{3}$$, then the value of $$y$$ is:

$$6(\sec^2 59^\circ - \cot^2 31^\circ) - \frac{2}{3}\sin 90^\circ - 3 \tan^2 56^\circ y \tan^2 34^\circ = \frac{y}{3}$$

$$6(\sec^2\ 59^{\circ}-\cot^2\left(90-59\right)^{\circ})-\frac{2}{3}\sin90^{\circ}-3\tan^2\ \left(90-34\right)^{\circ}\ y\ \tan^2\ 34^{\circ}=\frac{y}{3}$$

$$6(\sec^2\ 59^{\circ}-\tan^2\ 59^{\circ})-\frac{2}{3}\sin90^{\circ}-3\cot^2\ 34^{\circ}\ y\ \tan^2\ 34^{\circ}=\frac{y}{3}$$

We know that $$sin90^{\circ} = 1$$ and $$\sec^2\ \theta\ -\tan^2\ \theta\ \ =\ 1$$.

$$6\times\ 1-\frac{2}{3}\times\ 1-3y\times\ \frac{1}{\tan^2\ 34^{\circ}}\times\tan^2\ 34^{\circ}=\frac{y}{3}$$

$$6-\frac{2}{3}-3y=\frac{y}{3}$$

$$6-\frac{2}{3}=\frac{y}{3}+3y$$

$$\frac{18}{3}-\frac{2}{3}=\frac{y}{3}+\frac{9y}{3}$$

$$\frac{16}{3}=\frac{10y}{3}$$