Use implicit differentiation to find the equation of the line tangent to the graph of $3xy = y^2$ at $(1,3)$
This is what I tried so far: $\begin{align*} 3\frac{d}{dx}(x)\frac{d}{dx}(y(x)) &= \frac{d}{dx}(y(x)^2)\\ 3(1)\left(\frac{dy}{dx}\right) &= 2\frac{dy}{dx}\\ 3 &= \frac{dy}{dx} \end{align*}$
Then using that to solve the equation, I get: $\begin{align*} y-3 &= 3(x-1)\\ y &= 3x \end{align*}$
Did I solve this correctly?
EDIT:
Ok, so I would actually have: $\begin{align*} 3\left (\frac{dx}{dx}y+\frac{dy}{dx}(x)\right) &= 2y\frac{dy}{dx}\\ 3\left(y+x\left(\frac{dy}{dx}\right)\right) &= 2y\frac{dy}{dx}\\ 3\left(3+1\frac{dy}{dx}\right) &= 2(3)\frac{dy}{dx}\\ 9+3\frac{dy}{dx} &= 6\frac{dy}{dx}\\ 3 &= \frac{dy}{dx} \end{align*}$ Then using that to solve the equation, I get: $\begin{align*} y-3 &= 3(x-1)\\ y &= 3x \end{align*}$