Let $D=V(X_{0} X_{1}^{3} + X_{1}^{4} − X_{2}^{4})\subset\mathbb{P}_{\mathbb{C}}^{2}$ and $C=V(X_{0} X_{1}^{2} + X_{1}^{3} − X_{2}^{3})\subset\mathbb{P}_{\mathbb{C}}^{2}$. I have got that $C\cap D=\{(0:1:1), (1:-1:0), (1:0:0)\}$ with $mult_{(0:1:1)}(C,D)=1$, $mult_{(1:-1:0)}(C,D)=3$ and $mult_{(1:0:0)}(C,D)=8$. In addition, $C$ and $D$ have only one singular point, $(1:0:0)$, and only one inflection point, $(1:-1:0)$.
Now, I have to find a quartic $Q$ such that $mult_{(1:-1:0)}(Q,D)=3$, $mult_{(1:0:0)}(Q,D)=8$, and $(0:1:1), (0:1:0), (0:0:1), (1:0:1)\in Q$.
I think I have to manipulate $C$ in some way to get a quartic that satisfies those conditions, but I do not know how. In the same exercise I am also asked to find all lines tangent to $C$ and $D$ that contain $(0:0:1)$. I do not know if it is of any help, but (if I am not wrong) the only two are $V(X_{0}+X_{1})$ and $V(X_{1})$ for both curves, in the points $(1:-1:0)$ and $(1:0:0)$ respectively.
Any hint would be appreciated.
Let us write $F=X_{0} X_{1}^{3} + X_{1}^{4} − X_{2}^{4}$, $G=X_{0} X_{1}^{2} + X_{1}^{3} − X_{2}^{3}$ (which are the equations of $D$ and $C$ respectively).
Any quartic of the form $Q=V(GL)$, where $L$ is a linear polynomial, satisfies $$mult_{(0:1:1)}(Q,D)\ge mult_{(0:1:1)}(C,D)=1,$$ $$mult_{(1:-1:0)}(Q,D)\ge mult_{(1:-1:0)}(C,D)=3,$$ $$mult_{(1:0:0)}(C,D)\ge mult_{(1:0:0)}(C,D)=8.$$ And the same works if you choose $Q=V(GL+\lambda F)$, where $\lambda \in \mathbb{C}$. It suffices then to choose $L,\lambda$ such that $(0:1:0), (0:0:1), (1:0:1)\in Q$.
Since $F(0,1,0)=G(0,1,0)=1$, $F(0,0,1)=G(0,0,1)=-1$ and $F(1,0,1)=G(1,0,1)=-1$, you can take $\lambda=1$ and $L=X_1+X_2$.