[tex]\bf y=sin(3xy)\qquad \left(\frac{\pi }{6},1 \right)\\\\
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\cfrac{dy}{dx}=cos(3xy)\left[3\left[ y+x\frac{dy}{dx} \right] \right]\implies
\cfrac{dy}{dx}=cos(3xy)\left[3y+3x\frac{dy}{dx} \right]
\\\\\\
\cfrac{dy}{dx}=3ycos(3xy)+3xcos(3xy)\cfrac{dy}{dx}
\\\\\\
\cfrac{dy}{dx}-3xcos(3xy)\cfrac{dy}{dx}=3ycos(3xy)\impliedby \textit{common factor}[/tex]
[tex]\bf \cfrac{dy}{dx}[1-3xcos(3xy)]=3ycos(3xy)\implies \boxed{\cfrac{dy}{dx}=\cfrac{3ycos(3xy)}{1-3xcos(3xy)}}
\\\\\\
\left. \cfrac{3ycos(3xy)}{1-3xcos(3xy)} \right|_{\frac{\pi }{6},1}\implies \cfrac{3\cdot 1\cdot cos\left( 3\cdot \frac{\pi }{6}\cdot 1 \right)}{1-3\cdot \frac{\pi }{6}cos\left(3\cdot \frac{\pi }{6}\cdot 1 \right)}\\\\\\ \cfrac{3\cdot 1\cdot cos\left( \frac{\pi }{2} \right)}{1-\frac{\pi }{2}cos\left( \frac{\pi }{2} \right)}=0\\\\
-------------------------------\\\\[/tex]
[tex]\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=0(x-\frac{\pi }{6})\implies y-1=0
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
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\boxed{y=1}[/tex]
