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 $t$.
\[
x^{3}+4 t^{2}=65,2 y^{3}-3 t^{2}=80, t=4
\]
The slope of the curve at $t=4$ is (Type an integer or simplified fraction.)

Step 1 ：Given the equations \(x^{3}+4 t^{2}=65\) and \(2 y^{3}-3 t^{2}=80\), 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 \(\boxed{\frac{9}{10}}\).