= Solution
The <Hopf boundary point lemma> requires an <interior sphere condition>, strictness of the boundary maximum, and quantitative control of the <uniformly elliptic operator> near the contact point. Assume that $y\in\partial\Omega$, that $B_R(z)\subset\Omega$ and $\overline{B_R(z)}\setminus\{y\}\subset\Omega$, and that $|y-z|=R$. After shrinking the tangent <open ball>, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda\leq\Lambda$ and $B<\infty$,
$$
\lambda|\xi|^2\leq a^{ij}(x)\xi_i\xi_j\leq\Lambda|\xi|^2,
\qquad |b(x)|\leq B.
$$
The coefficient <matrix> may be replaced by its <symmetric matrix> part because the <Hessian matrix> is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent <open ball>. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is <connected>, the <strong maximum principle for elliptic operators> supplies this strictness from a nonconstant solution attaining its global maximum at $y$.
Put $\mathbf n=(y-z)/R$, the outward <unit normal> of the tangent <open ball>. With only the given <continuity> at $y$, the precise conclusion is
$$
\boxed{\liminf_{t\downarrow0}\frac{u(y)-u(y-t\mathbf n)}{t}>0.}
$$
\b[If the outward normal derivative exists, it is strictly positive.] In particular, if $u$ is <continuously differentiable> up to $y$ and the domain has this outward <unit normal>, then $\partial_{\mathbf n}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a <normal derivative>.
For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the <barrier for the Dirichlet problem>
$$
v(x)=e^{-k|x-z|^2}-e^{-kR^2}.
$$
Writing $q=x-z$, direct <differentiation> gives
$$
Lv=e^{-k|q|^2}\bigl(4k^2a^{ij}q_iq_j-2k\operatorname{tr}a-2kb\cdot q\bigr)
\geq e^{-k|q|^2}\bigl(k^2\lambda R^2-2k(n\Lambda+BR)\bigr).
$$
Choose $k>2(n\Lambda+BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, <compactness> and the strict maximum give $m=\min_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\varepsilon>0$ with $\varepsilon\max_{|x-z|=R/2}v\leq m$. On the outer sphere $v=0$ and $u-u(y)\leq0$, including at $y$ by <continuity>. Consequently $w=u-u(y)+\varepsilon v$ has $Lw\geq0$ and nonpositive boundary values. The allowed comparison principle, or the <weak maximum principle for elliptic operators>, yields $w\leq0$ in $A$.
Along the inward radius this gives
$$
\frac{u(y)-u(y-t\mathbf n)}t\geq
\varepsilon\frac{e^{-k(R-t)^2}-e^{-kR^2}}t
\longrightarrow 2\varepsilon kR e^{-kR^2}>0.
$$
This proves the <Hopf boundary point lemma>. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.
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