After reading your last comment, I thought it might help to elaborate on my comment in an answer.
The elementary functions are holomorphic because exponentials and powers and their inverses are holomorphic, constants are holomorphic, arithmetic combinations (sums, products, etc.) of holomorphic functions are holomorphic, and compositions of holomorphic functions are holomorphic.
The function $f(z)=\overline{z}$ is not holomorphic because its complex derivative does not exist at any point. If $z\neq w$, then $\displaystyle{\frac{f(z)-f(w)}{z-w}=\frac{\overline{z-w}}{z-w}=\exp(-2i\mathrm{arg}(z-w))}$. You can have $z$ arbitrarily close to $w$ while $\exp(-2i\mathrm{arg}(z-w))$ can be an arbitrary point on the unit circle. To see this, you can consider $z$ moving in tiny circles around $w$. To see more explicitly what can go wrong, consider $z$ approaching $w$ along the horizontal line $z=w+t$, where $t$ is real and approaches zero, and then consider $z$ approaching $w$ along the vertical line $z=w+it$, where again $t$ is real and approaches $0$. In the first case the quotient is always $1$, and in the second case the quotient is always $-1$. Therefore $\displaystyle{\lim_{z\to w}\frac{f(z)-f(w)}{z-w}}$ does not exist.
The Cauchy-Riemann equations give a less direct (but perhaps easier) way to see that $f$ is not holomorphic. If $g$ is a holomorphic function, and if $g$ is expressed as a function of $x$ and $y$ (with a little abuse of notation) $g(x,y)=g(x+iy)$, then $\displaystyle{i\frac{\partial g}{\partial x}=\frac{\partial g}{\partial y}}$. In the case of $f(z)=\overline{z}$, you have $f(x,y)=x-iy$, so $\displaystyle{i\frac{\partial f}{\partial x}=i}$ and $\displaystyle{\frac{\partial f}{\partial y}=-i}$, showing that the Cauchy-Riemann equations are not satisfied.