[tex]2x-3y=6\\ \\ 3y=2x-6\\ \\ \frac { 1 }{ 3 } \cdot 3y=\frac { 1 }{ 3 } \left( 2x-6 \right) \\ \\ y=\frac { 2 }{ 3 } x-\frac { 6 }{ 3 } \\ \\ y=\frac { 2 }{ 3 } x-2\\ \\ \therefore \quad \frac { dy }{ dx } =\frac { 2 }{ 3 } [/tex]
This is because:
[tex]y=k{ x }^{ n }\\ \\ \frac { dy }{ dx } =kn{ x }^{ n-1 }[/tex]
And also because when you differentiate a constant (i.e the number 2), what you always get is the value 0.
