To find the equations of the tangent and normal lines to the curve at point $P(1,1)$, given by the equation:

$x^3 y - 2xy^2 + x^2 - 3y + 3 = 0$

we'll proceed with these steps:

Step 1: Differentiate the Equation Implicitly

We need to implicitly differentiate the equation with respect to $x$ to find the slope of the tangent line.

Given: $f(x, y) = x^3 y - 2xy^2 + x^2 - 3y + 3 = 0$

Differentiating $f(x, y) = 0$ with respect to $x$: $\frac{d}{dx}(x^3 y) - \frac{d}{dx}(2xy^2) + \frac{d}{dx}(x^2) - \frac{d}{dx}(3y) + \frac{d}{dx}(3) = 0$

Applying the product rule to each term:

1. $\frac{d}{dx}(x^3 y) = 3x^2 y + x^3 \frac{dy}{dx}$
2. $\frac{d}{dx}(2xy^2) = 2y^2 + 2x(2y \frac{dy}{dx})$
3. $\frac{d}{dx}(x^2) = 2x$
4. $\frac{d}{dx}(3y) = 3 \frac{dy}{dx}$
5. $\frac{d}{dx}(3) = 0$

So we have: $3x^2 y + x^3 \frac{dy}{dx} - 2y^2 - 4xy \frac{dy}{dx} + 2x - 3 \frac{dy}{dx} = 0$

Step 2: Substitute Point $P(1,1)$ and Solve for $\frac{dy}{dx}$

At $P(1,1)$, substitute $x = 1$ and $y = 1$:

1. $3(1)^2(1) = 3$
2. $(1)^3 \frac{dy}{dx} = \frac{dy}{dx}$
3. $- 2(1)^2 = -2$
4. $- 4(1)(1) \frac{dy}{dx} = -4 \frac{dy}{dx}$
5. $2(1) = 2$
6. $- 3 \frac{dy}{dx} = -3 \frac{dy}{dx}$

Substitute these values into the differentiated equation: $3 + \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$

Combine terms: $3 - 2 + 2 + (1 - 4 - 3) \frac{dy}{dx} = 0$

Simplify: $3 - 4 \frac{dy}{dx} = 0$

Thus: [ \frac{dy}{dx}