Find \( \frac{d y}{d x} \) at the given point. \[ 3 x e^{7 y}=3 \text { at }(1,0) \]

To find \( \frac{d y}{d x} \) for the equation \( 3 x e^{7 y} = 3 \) at the point \( (1, 0) \), we'll use implicit differentiation. 1. Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(3 x e^{7 y}) = \frac{d}{dx}(3) \] 2. Using the product rule on the left side: \[ 3(e^{7y} + x \frac{d}{dx}(e^{7y})) = 0 \] 3. For \( \frac{d}{dx}(e^{7y}) \), we use the chain rule: \[ \frac{d}{dx}(e^{7y}) = e^{7y} \cdot \frac{d(7y)}{dx} = 7 e^{7y} \frac{dy}{dx} \] 4. Substitute this back: \[ 3 \left( e^{7y} + 7x e^{7y} \frac{dy}{dx} \right) = 0 \] 5. Distributing the 3 gives: \[ 3 e^{7y} + 21 x e^{7y} \frac{dy}{dx} = 0 \] 6. Solving for \( \frac{dy}{dx} \): \[ 21 x e^{7y} \frac{dy}{dx} = -3 e^{7y} \] \[ \frac{dy}{dx} = -\frac{3 e^{7y}}{21 x e^{7y}} = -\frac{3}{21x} = -\frac{1}{7x} \] 7. Now, substitute \( x = 1 \) and \( y = 0 \): \[ \frac{dy}{dx} \bigg|_{(1, 0)} = -\frac{1}{7 \cdot 1} = -\frac{1}{7} \] Thus, \( \frac{d y}{d x} \) at the point \( (1, 0) \) is \(-\frac{1}{7}\).