To solve this problem, we are given:

$\ln(5y) \cos x = y^6 + 14xy$

We need to find $y'$ or $\frac{dy}{dx}$. Let's proceed by differentiating both sides with respect to $x$, applying implicit differentiation.

Step-by-Step Solution:

1. Differentiate the Left Side:
   • The left side is $\ln(5y) \cos x$.
   • Using the product rule for differentiation, we get: $\frac{d}{dx} \left( \ln(5y) \cos x \right) = \frac{d}{dx} (\ln(5y)) \cdot \cos x + \ln(5y) \cdot \frac{d}{dx} (\cos x)$
   • Differentiate $\ln(5y)$ with respect to $x$: since $\ln(5y)$ is $\frac{1}{5y} \cdot 5 = \frac{1}{y}$, and by the chain rule, it becomes $\frac{1}{y} \cdot y'$.
   • This gives: [ y'