Step 1: Recognize the AM-GM pattern.
In the equation $ \frac{y}{x}\cos^4\alpha+\frac{x}{y}\sin^4\alpha=2\sin^2\alpha\cos^2\alpha $, note the RHS equals $ 2\sqrt{\frac{y}{x}\cos^4\alpha\cdot\frac{x}{y}\sin^4\alpha} $. So LHS = $ 2\times $geometric mean of the two terms.
Step 2: Apply AM-GM equality condition.
Since $ P+Q=2\sqrt{PQ} $ holds iff $ P=Q $: \[ \frac{y}{x}\cos^4\alpha = \frac{x}{y}\sin^4\alpha \]
Step 3: Solve for $ y/x $.
\[ y^2\cos^4\alpha = x^2\sin^4\alpha \implies \frac{y}{x} = \tan^2\alpha \]
Step 4: Write the explicit relation.
\[ y = x\tan^2\alpha \] where $ \tan^2\alpha $ is a constant ($ \alpha $ is a fixed parameter).
Step 5: Differentiate.
\[ \frac{dy}{dx} = \tan^2\alpha \]
Step 6: Express in the required trigonometric form.
\[ \frac{dy}{dx} = \frac{\sin^2\alpha}{\cos^2\alpha} \] \[ \boxed{\dfrac{\sin^2\alpha}{\cos^2\alpha}} \]