Find $\frac{d y}{d t}$ for the equation below. \[ 5 x^{3}+4 x y+y^{4}=3 \]

Step 1 ：We are given the equation \(5x^{3} + 4xy + y^{4} = 3\). We are asked to find \(\frac{dy}{dt}\).

Step 2 ：We start by differentiating both sides of the equation with respect to \(t\). We will use the chain rule and the product rule.

Step 3 ：Applying the product rule to \(4xy\), we get \(4x\frac{dy}{dt} + 4y\frac{dx}{dt}\).

Step 4 ：Applying the chain rule to \(y^{4}\), we get \(4y^{3}\frac{dy}{dt}\).

Step 5 ：Applying the chain rule to \(5x^{3}\), we get \(15x^{2}\frac{dx}{dt}\).

Step 6 ：Putting it all together, we get \(15x^{2}\frac{dx}{dt} + 4x\frac{dy}{dt} + 4y\frac{dx}{dt} + 4y^{3}\frac{dy}{dt} = 0\).

Step 7 ：We can simplify this to \(\frac{dy}{dt} = \frac{-15x^{2}\frac{dx}{dt} - 4y\frac{dx}{dt}}{4x + 4y^{3}}\).

Step 8 ：However, this is not the final answer because it contains \(\frac{dx}{dt}\), which is not given in the problem. We need to simplify the expression by assuming \(\frac{dx}{dt} = 1\).

Step 9 ：Substituting \(\frac{dx}{dt} = 1\) into the equation, we get \(\frac{dy}{dt} = \frac{-15x^{2} - 4y}{4x + 4y^{3}}\).

Step 10 ：Final Answer: The derivative of \(y\) with respect to \(t\) is \(\boxed{\frac{-15x^{2} - 4y}{4(x+y^{3})}}\).