How do you find $\frac{dH}{dt}$ where,
$$H(x,y)=\sin(3x-y), x^3+2y=2t^3, x-y^2=t^2+3t$$
I think I'm supposed to use the multivariable chain rule,
$$\frac{dH}{dt}=\frac{\partial H}{\partial x}\frac{dx}{dt}+\frac{\partial H}{\partial y}\frac{dy}{dt}$$
$$\frac{\partial H}{\partial x}=3\cos(3x-y), \frac{\partial H}{\partial y}=-\cos(3x-y)$$
So factoring out $cos(3x-y)$ I have,
$$\frac{dH}{dt}=\cos(3x-y)[3\frac{dx}{dt}-\frac{dy}{dt}]$$
The problem is that I have no idea how to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ from the above equations.
The correct answer is $$\frac{dH}{dt}=\frac{36t^2y+12t+9x^2-6t^2+6x^2t+18}{6x^2y+2}\,\cos(3x-y)$$
Just differentiate wrt to $t$ $$ \begin{cases} x^3+2y=2t^3 \\ x-y^2=t^2+3t \end{cases} $$ $$ \implies \begin{cases} 3x^2x'+2y'=6t^2 \\ x'-2yy'=2t+3 \end{cases} $$
where $x'=\frac {dx}{dt}$ and $y'=\frac {dy}{dt}$
Solve the system to express $x'$ and $y'$ as function of $x,y$