= Solution
There is a <nonsingular point of an algebraic curve> on $C$, since part (b) makes its <singular locus> finite whereas a <projective curve> has infinitely many complex points. Apply a <projective transformation> taking one such point to $p=[1:0:0]$, and use $P$ for the transformed <homogeneous polynomial>. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common <irreducible component> with $C$.
The <Euler theorem for homogeneous functions> and its differentiated identities give
$$
P_x(p)=dP(p)=0,
\qquad
P_{xx}(p)=(d-1)P_x(p)=0,
$$
and
$$
P_{xy}(p)=(d-1)P_y(p),
\qquad
P_{xz}(p)=(d-1)P_z(p).
$$
Since $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the <polar curve> $D$ is also nonsingular at $p$, and these identities show that $T_pD=T_pC$. Hence $I_p(C,D)\geq2$.
This additional intersection is separate from all the <singular points> of $C$. Thus <Bézout's theorem> gives
$$
2s+2\leq\sum_{q\in C\cap D}I_q(C,D)\leq d(d-1),
\qquad
\boxed{s\leq\frac{d(d-1)}2-1.}
$$
The improvement comes from making the <polar curve> tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.
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