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(x2+y2–1)3–x2y3=0 canonical form

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-3 + 3*x2 + 3*y2 - x2*y3 = 0
$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$
-x2*y3 + 3*x2 + 3*y2 - 3 = 0
Invariants method
Given equation of the surface of 2-order:
$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$
This equation looks like:
$$a_{11} y_{3}^{2} + 2 a_{12} y_{2} y_{3} + 2 a_{13} x_{2} y_{3} + 2 a_{14} y_{3} + a_{22} y_{2}^{2} + 2 a_{23} x_{2} y_{2} + 2 a_{24} y_{2} + a_{33} x_{2}^{2} + 2 a_{34} x_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 0$$
$$a_{34} = \frac{3}{2}$$
$$a_{44} = -3$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

substitute coefficients
$$I_{1} = 0$$
     |0  0|   |0  0|   | 0    -1/2|
I2 = |    | + |    | + |          |
     |0  0|   |0  0|   |-1/2   0  |

$$I_{3} = \left|\begin{matrix}0 & 0 & - \frac{1}{2}\\0 & 0 & 0\\- \frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & - \frac{1}{2} & 0\\0 & 0 & 0 & \frac{3}{2}\\- \frac{1}{2} & 0 & 0 & \frac{3}{2}\\0 & \frac{3}{2} & \frac{3}{2} & -3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & - \frac{1}{2}\\0 & - \lambda & 0\\- \frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
     |0  0 |   | 0   3/2|   | 0   3/2|
K2 = |     | + |        | + |        |
     |0  -3|   |3/2  -3 |   |3/2  -3 |

     |0   0    0 |   | 0    0   3/2|   | 0    -1/2   0 |
     |           |   |             |   |               |
K3 = |0   0   3/2| + | 0    0   3/2| + |-1/2   0    3/2|
     |           |   |             |   |               |
     |0  3/2  -3 |   |3/2  3/2  -3 |   | 0    3/2   -3 |

$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{9}{2}$$
$$K_{3} = \frac{3}{4}$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
Solve this equation
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde y2^{2} \lambda_{2} + \tilde y3^{2} \lambda_{1}\right) = 0$$
$$3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
and
$$- 3 \tilde x2 + \frac{\tilde y2^{2}}{2} - \frac{\tilde y3^{2}}{2} = 0$$
$$- 2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
and
$$2 \tilde x2 - \left(\frac{\tilde y2^{2}}{3} - \frac{\tilde y3^{2}}{3}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form