The question states to find the equation of the tangent and normal to the curve $y^3-2y-x^3+x=15$ at (2,3).
What I did was to find the gradient of that equation and ended up with $\frac{dy}{dx}-\frac{dy}{dx}=\frac{\left(3x^2-1\right)}{2y^2}$ Not sure if I am correct but I am stuck at this point.
$y^3-2y-x^3+x-15=0$
Differentiating the equation wrt $x$, we get
$3y^2\frac{dy}{dx}-2\frac{dy}{dx}-3x^2+1=0$
Therefore, $\frac{dy}{dx}=\frac{3x^2-1}{3y^2-2}$
Put $x=2$ and $y=3$ to get the slope of tangent.
I suppose you can take it from here and get the required equations of tangent and normal.