I got it.

First the differentiation:

$\frac{d}{dx}x^2+3 y(x)^2 = \frac{d}{dx}3xy(x)+1$

Differentiate the sum term by term and factor out constants:

$\frac{d}{dx}x^2+3 \frac{d}{dx}y(x)^2 = \frac{d}{dx}3xy(x)+1$

The derivative of $x^2$ is $2x$:

$3\frac{d}{dx}y(x)^2+2x = \frac{d}{dx}3xy(x)+1$

Use the chain rule:

$\frac{d}{dx}y(x)^2 = \frac{du^2}{du} \frac{du}{dx}$
where $u = y(x)$ and $\frac{du^2}{du} = 2 u$:

$6y(x) \frac{d}{dx}y(x)+2x = \frac{d}{dx}3xy(x)+1$

The derivative of $y(x)$ is $y'(x)$:

$6y(x)y'(x)+2x = \frac{d}{dx}3xy(x)+1$

Differentiate the sum term by term and factor out constants:

$6y(x) y'(x)+2x = 3 \frac{d}{dx}xy(x)+\frac{d}{dx(1)}$

The derivative of 1 is zero:

$6y(x) y'(x)+2x = 3 \frac{d}{dx}xy(x)$

Use the product rule,
$\frac{d}{dx(u v)} = v \frac{du}{dx}+u\frac{dv}{dx}$
where $u = x$ and $v = y(x)$:

$6 y(x)y'(x)+2x = 3 (x \frac{d}{dx}y(x)+y(x) \frac{d}{dx(x)})$

The derivative of $y(x)$ is $y'(x)$:

$6 y(x)y'(x)+2x = 3y(x) \frac{d}{dx(x)}+3xy'(x)$

The derivative of $x$ is 1:

$6 y(x)y'(x)+2x = 3xy'(x)+3y(x)$

To find where $x=1$ just put $x=1$ into the original equation to get the y equation $y^2+y=0$ where the roots are $y=1$ or $y=0$

The tangent is vertical where $\frac{dy}{dx}=$undefined (or infinite)