Find the equation of the tangent to the curve

Find the equation of the tangent to the curve $2x^{3}-x^{2}y^{2}-3x+y+1=0$ at the point where $y=1$.

Step 1: Where does $y=1$ meet the curve?

Substituting $y=1$: \[2x^{3}-x^{2}-3x+2 = 0. \tag{1}\]

Testing $x=1$: $2-1-3+2=0\ \checkmark$; dividing out $(x-1)$, \[2x^{3}-x^{2}-3x+2 = (x-1)(2x^{2}+x-2). \tag{2}\]

The quadratic $2x^{2}+x-2$ has discriminant $\Delta=1+16=17>0$, giving two further real (irrational) roots $x=\dfrac{-1\pm\sqrt{17}}{4}$. So the line $y=1$ actually meets the curve at three points, not one — the problem's phrase “the point where $y=1$” is only unambiguous once we notice that a single one of the three, $x_{0}=1$, gives integer data and is evidently the point intended.

Step 2: Implicit differentiation

With $F(x,y)=2x^{3}-x^{2}y^{2}-3x+y+1$, differentiating $F=0$: \[6x^{2}-2xy^{2}-2x^{2}y\,y'-3+y' = 0 \;\Longrightarrow\; y' = \frac{-6x^{2}+2xy^{2}+3}{1-2x^{2}y}. \tag{3}\]

Step 3: Slope at the intended point $P=(1,1)$

\[ y'(1,1) = \frac{-6+2+3}{1-2} = \frac{-1}{-1} = 1. \tag{4} \]

Step 4: Equation of the tangent

\[ y-1 = 1\cdot(x-1) \;\Longrightarrow\; y=x. \tag{5} \]

Step 5: Verification

\[(1,1) \text{ on the curve: } 2-1-3+1+1=0\ \checkmark\]

\[\text{Slope: numerator } -6+2+3=-1,\ \text{denominator } 1-2=-1,\ \text{ratio } 1\ \checkmark\]

\[\text{Tangent at }(1,1)\text{: } y(1)=1\ \checkmark\]