Use implicit differentiation to find $\frac{d y}{d x}$. \[ e^{3 x y}=7 y \] \[ \frac{d y}{d x}=\square \]

Step 1 ：Differentiate both sides of the equation \(e^{3xy} = 7y\) with respect to \(x\)

Step 2 ：Apply the chain rule to the left side: \(\frac{d}{dx}(e^{3xy}) = 3y \cdot e^{3xy} \cdot \frac{dx}{dx} + 3x \cdot e^{3xy} \cdot \frac{dy}{dx}\)

Step 3 ：Differentiate the right side: \(\frac{d}{dx}(7y) = 7 \cdot \frac{dy}{dx}\)

Step 4 ：Combine the derivatives: \(3y \cdot e^{3xy} + 3x \cdot e^{3xy} \cdot \frac{dy}{dx} = 7 \cdot \frac{dy}{dx}\)

Step 5 ：Isolate terms with \(\frac{dy}{dx}\): \(3x \cdot e^{3xy} \cdot \frac{dy}{dx} - 7 \cdot \frac{dy}{dx} = -3y \cdot e^{3xy}\)

Step 6 ：Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(3x \cdot e^{3xy} - 7) = -3y \cdot e^{3xy}\)

Step 7 ：Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-3y \cdot e^{3xy}}{3x \cdot e^{3xy} - 7}\)

Step 8 ：\(\boxed{\frac{dy}{dx} = \frac{-3y \cdot e^{3xy}}{3x \cdot e^{3xy} - 7}}\)