You will be an expert high school math teacher. Explain the worked solution to the following problem step-by-step, according to the NAME_1 scheme I also provide. In the NAME_1 scheme I also give information about the type of NAME_1 rewarded as follows:
• M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise indicated.
• A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
• B marks are unconditional accuracy marks (independent of M marks). 
These letters are wrapped in [ ] e.g. [M1] indicates 1 NAME_1 for method being awarded.
• d means dependent.

Question: The curve C has equation $xy^2 = x^2y+6; x\neq0,y\neq0$
Find an equation for the tangent to C at the point P (2, 3), giving your answer in the form
ax + by + c = 0 where a, b and c are integers.

NAME_1 scheme:
$xy^2 --> (...+)2xy\frac{dy}{dx}$ or $x^2y --> (...+)x^2 \frac{dy}{dx}$ but see notes. [M1] 

$xy^2 --> y^2+kxy\frac{dy}{dx}$ or $x^2y-->kxy+x^2\frac{dy}{dx}$. [dM1]

$y^2+2xy\frac{dy}{dx} = 2xy+x^2\frac{dy}{dx}$ or equivalent. [A1]

$(x^2-2xy)\frac{dy}{dx}=y^2-2xy-->\frac{dy}{dx}=\frac{y^2-2xy}{x^2-2xy}-->\frac{dy}{dx}=\frac{3^2-12}{2^2-12}=...$ [M1]

Tangent is $y-3=\frac{3}{8}(x-2)$. [dM1]

$3x-8y+18=0$ [A1] Do not say the word "equation" in the response
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