A curve that intersects $2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ with multiplicity $3$ or more let $C=2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ be a curve in $\mathbb{C^2}$. Find a conic that intersects $C$ in $(2,2)$ with intersection multiplicity $\geq 3$.
Since $C_x(2,2) \neq 0$ and $C_y(2,2) \neq 0$ it is a nonsingular point for $C$. We have defined the multiplicity of intersection of two curves in the projective plane so I imagine I need to find a conic $D=a_{00}x^2+2a_{01}xy+2a_{02}xz+a_{11}y^2+2a_{12}yz+a_{22}z^2$ such that eliminating $z$ from $D$ and $C^h$, where $C^h$ is the homogenization of $C$, the factor $(x-y)^3$ divides such polynomial.
I computed such conditions but they are quite unusable in this form so I opted for a trial-and-error approach, but all of the conics I tried have i.m. $2$, like the circle $(x-\frac{7}{4})^2+(y-3)^2=\frac{17}{16}$.
Any idea or hint? Thank you.

If it is not required the conic should be smooth, you could take the union of the tangent to $C$ at $(2,2)$ and any other line that passes through $(2,2)$.
