[tex]\text{Given that,}\\\\~~~1+ \ln (3xy) = e^{3x-y}\\\\\\\implies \dfrac d{dx} \left[1 + \ln (3xy) \right] = \dfrac{d}{dx} \left(e^{3x-y} \right)\\\\\\\implies \dfrac{1}{3xy} \cdot 3 \dfrac{d}{dx}(xy) = e^{3x-y} \cdot \left(3 - \dfrac{dy}{dx} \right)\\ \\\\\implies \dfrac 1{xy} \left( x \dfrac{dy}{dx} + y \right)= e^{3x-y} \cdot \left(3 - \dfrac{dy}{dx} \right)\\\\\\\implies \dfrac 1y \dfrac{dy}{dx} + \dfrac 1x = 3e^{3x-y}- e^{3x -y} \dfrac{dy}{dx}[/tex]
[tex]\implies \dfrac 1y \dfrac{dy}{dx} + e^{3x-y} \dfrac{dy}{dx} = 3e^{3x-y} - \dfrac 1x\\\\\\\implies \left( \dfrac 1y+ e^{3x-y} \right)\dfrac{dy}{dx} = 3e^{3x-y} - \dfrac 1x\\\\\\\implies \dfrac{dy}{dx}=\dfrac{3 e^{3x-y} -\dfrac 1x}{\dfrac 1y+ e^{3x-y} }\\\\\text{At point}~ (x,y) = (1/3, 1), \text{slope of the tangent line,}\\\\m=\dfrac{dy}{dx} = 0\\\\\text{Hence the equation of tangent line,}\\\\~~~y-1 =0\\\\\implies y=1[/tex]
