Let's cast the complex function $f(z) = u(z) + iv(z), z = x+iy$, as the multivariable function $F(x,y) = U(x,y) + iV(x,y) ; x,y \in R$.
Thus, $$dF = F_x\,dx + F_y\,dy = U_x\,dx + iV_x\,dx + U_y dy + iV_y\,dy$$
Making use of $$dz = dx + i\,dy,$$ $$dz∗ =dx−i\,dy,$$
We plug in $$dx = \frac 1 2 (dz + dz∗),$$ $$dy= \frac 1 {2i}(dz−dz∗)$$
into the equation of $dF$, thus we get:
$$dF = \frac 1 2 (U_x + V_y + i(V_x - U_y))\, dz + \frac 1 2 (U_x - V_y + i(V_x + U_y))\, dz*$$
Now it can be seen that if the function $f(z)$ (and thus $F(x,y)$) satisfies the Cauchy Riemann Equations, then the factor in front of $dz*$ will be zero, and we will have the expression for the derivative of the function $F(x,y)$ (and hence $f(z)$).
Now there are two assumptions for the derivative of a complex function to exist, namely that (1) it satisfies CR equations, and (2) it has continuous first partial derivatives.
But we just found the derivative of $f(z)$, without assuming the condition of continuous first partials!! How is this?
The exterior derivative $dF$ isn't really well-defined if we just assume existence of partial derivatives. Of course, we could formally write: $$ dF = F'_x\,dx + F'_y\,dy $$ but if we don't know that $F$ is differentiable, $dF$ doesn't have very nice properties. (For example, invariance under coordinate transformations is no longer guaranteed.) In particular, just because you can read off that $\frac{\partial F}{\partial \bar z} = 0$, it doesn't imply that $f'$ exists.
As a concrete counterexample, let $$ F(z) = \begin{cases} \exp(-1/z^4), & z \neq 0 \\ 0, & z = 0 \end{cases} $$ You can check that $F$ has partial derivatives everywhere and that they satisfy Cauchy-Riemann's equations, but of course $F$ is not even continuous at $z=0$.
On a related, but different note. If $F$ is continuous and has partial derivatives which satisfy Cauchy-Riemman's equations, then $F$ is in fact holormorphic. This result is known as Looman-Menchoff's theorem and is pretty difficult.