Assuming that the equation defines $x$ and $y$ implicitly as differentiable functions $x=f(t),y=g(t)$, find the slope of the curve $x=f(t)$, $y=g(t)$ at the given value of $t$.
$$x^3+4t^2=65,2y^3-3t^2=80,t=4$$
The slope of the curve at $t=4$ is (Type an integer or simplified fraction.)

Step 1 ：Given the equations $x^3+4t^2=65$ and $2y^3-3t^2=80$, we are asked to find the slope of the curve at $t=4$.

Step 2 ：We start by finding the derivative of $x=f(t)$ and $y=g(t)$ with respect to $t$.

Step 3 ：Using the chain rule, we differentiate both sides of the equation with respect to $t$ to get the derivative of $x$ and $y$ with respect to $t$.

Step 4 ：For $x=f(t)$, the derivative $dx/dt$ is $3t^2+8t$.

Step 5 ：For $y=g(t)$, the derivative $dy/dt$ is $6t^2-6t$.

Step 6 ：We then substitute $t=4$ into the derivatives to find the slope of the curve at $t=4$.

Step 7 ：The slope of the curve at $t=4$ is $\frac{9}{10}$.

Step 8 ：Final Answer: The slope of the curve at $t=4$ is $\boxed{{\displaystyle \frac{9}{10}}}$.