Need to find the slope of the tangent line at (2/3, 4/3). Since the original equation cannot be easily solved for y, use implicit differentiation. The understanding is that y is some function of x, whenever we take the derivative of y it is with respect to x, i.e. dy/dx. In the following y' is used instead of dy/dx.
On the left hand side use the product rule where the "first function is 3x" and the second is y, at least that is one way to think about it.
3xy' + 3y = 3x2 + 3y2y'
Notice at this point we can divide all terms by 3 as part of the simplification. Transposing terms containing y' to left side and others to right side we have:
xy' - y2y' = x2 - y
Factor out the y' from both terms on the left. Divide the right side by the coefficient of y'. Now use (2/3, 4/3) in the formula for y' to get the slope of the tangent line at that point (4/5). Now use point-slope to write the equation.
Here is a picture: