conjugate of conjugate is continuous

If $\gamma:[a,b] \rightarrow \mathbb{C}$ is a piecewise smooth differentiable curve. Let $\overline{\gamma}$ denote the image of $\gamma$ under $z \mapsto \overline{z}$. Show if $f$ is continuous on $\gamma$, then $z \mapsto \overline{f(\overline{z})}$ is continuous and

$$\overline{\int_\gamma f(z) dz}=\int_\overline{\gamma} \overline{f(\overline{z})} dz$$

So from my understanding, $z \mapsto \overline{z}$ is the map sending $x+iy \mapsto x-iy$. How do I conjugate the whole function and the variable $z$? To show the map is continuous, is it just a limit argument? Any hints greatly appreciated! Could I let $f(x,y)=u(x,y)+iv(x,y)$ where $u,v$ satisfy Cauchy Riemann?

Let $c(z)=\overline z$. Then $c$ is continuous and $\overline{f\left(\overline z\right)}$ is just $(c\circ f\circ c)(z)$. Since $z\mapsto\overline{f\left(\overline z\right)}$ is the composition of three continuous maps, it is continuous.

On the other hand,$$\overline{\int_\gamma f(z)}\,\mathrm dz=\overline{\int_a^bf\bigl(\gamma(t)\bigr)\gamma'(t)}\,\mathrm dt\tag1$$and\begin{align}\int_{\overline\gamma}\overline{f\left(\overline z\right)}\,\mathrm dz=\int_a^b\overline{f\bigl(\gamma(t)\bigr)}\overline\gamma'(t)\,\mathrm dt\tag2\end{align}and it is clear that $(1)=(2)$.

Write $\int_{\gamma} f(z)\, dz = \int_a^b f(\gamma(t))\,\gamma'(t)\, dt$. The conjugate of this integral is $$\int_a^b \overline{f(\gamma(t))}\, \overline{\gamma'(t)}\, dt = \int_a^b \overline{f(\gamma(t))}\, \overline{\gamma}'(t)\, dt$$ Since $t\mapsto \overline{\gamma}(t)$ parametrizes $\overline{\gamma}$, the latter integral is $\int_{\overline{\gamma}} \overline{f(\bar{z})}\, dz$.