Let
$$f(z)=\begin{cases} \frac{\bar{z}^2}{z} &\text{ if } z\ne 0\\ 0 &\text{ if } z=0 \end{cases}$$
We can show that Cauchy-Riemann equations are satisfied at $(0,0)$, but we can also show that $f$ is not differentiable at $0$. Is this because $f$ is discontinuous at $0$? Just want to make sure I'm on the right way.
The function $$f(z)=\begin{cases} \frac{\overline{z}^2}{z} &\text{ if } z\ne 0\\ 0 &\text{ if } z=0 \end{cases}$$ does indeed satisfy the Cauchy-Riemann equations but is not complex differentiable in $0$ because it is not real-differentiable in $(0,0)$, which is the other condition needed to use the Cauchy-Riemann criterion. This however can not be seen by a discontinuity of $f$, because $f$ is continuous in $(0,0)$. We have $|f(z)-f(0)| = |z| \to 0\quad(z\to 0)$.
We can use an other argument: If $f$ was complex differentiable then we would have $$f'(0) = \lim_{z\to 0}\frac{f(z)-f(0)}{z} = \lim_{z\to 0}\frac{\bar z^2}{z^2}.$$ This limit doesnt exist because if we plug in the sequence $(\frac 1n)_{n\in\mathbb N}$, then the limit evaluates to $1$ and if we plug in the sequence $(\frac{1}{n}+i\frac 1n)_{n\in\mathbb N}$, we get $$\frac{(\overline{\frac 1n+i\frac 1n})^2}{(\frac 1n+i\frac 1n)^2} = \frac{(\frac 1n-i\frac 1n)^2}{(\frac 1n+i\frac 1n)^2} = \frac{\frac{1}{n^2}-\frac{2i}{n^2}-\frac{1}{n^2}}{\frac{1}{n^2}+\frac{2i}{n^2}-\frac{1}{n^2}} = -1$$ for all $n\in\mathbb N$, hence also in the limit, which doesn't coincide with the other limit.
Original answer: (this refered to an old version of the quesion where $f(z) = \frac{\bar z}{z}$ for $z\neq 0$.)
In the following I will write $f(x,y)$ for $f(x+iy)$.
You can impossibly have shown that $f$ satisfies the Cauchy-Riemann equations in $(0,0)$ because $f$ is not partially differentiable at this point.1 Note that $$\frac{\partial f}{\partial x}(0,0) := \lim_{h\to 0} \frac{f((0,0)+h(1,0))-f(0,0)}{h}.$$ In general, this is not the same as calculating $\frac{\partial f}{\partial x}(x,y)$ outside of $(0,0)$ and then plugging in $(0,0)$ in this solution which is not made for this specific point. (This is probably what you have done and how you came to the conclusion that $f$ satisfies the Cauchy-Riemann equations?)
In our case we have $$\lim_{h\to 0} \frac{f((0,0)+h(1,0))-f(0,0)}{h} = \lim_{h\to0}\frac{f(h(1,0))}{h} = \lim_{h\to0}\frac{1}{h}\frac{\bar h}{h} = \lim_{h\to0}\frac{1}{h}\frac{h}{h} = \lim_{h\to0}\frac{1}{h}$$ which obviously doesn't exist. The same goes for $\frac{\partial f}{\partial y}(0,0)$.
1If $f$ isn't partially differentiable at $(0,0)$ at least one of $\operatorname{Re}f, \operatorname{Im}f$ isn't, hence $f$ can't satisfy the Cauchy-Riemann equations.
As you said, the problem is that there is a discontinuity in $z=0$. To check it, let's pick two sequences which converge to zero when $n$ tends to infinity. For example,
$$\{z_n\}_{n=1}^\infty= \left\{\frac{1}{n}\right\}_{n=1}^\infty \qquad \text{and} \qquad \{w_n\}_{n=1}^\infty= \left\{\frac{i}{n}\right\}_{n=1}^\infty .$$
We know that $f$ is continuous in $z_0$ iff for all sequence $\{z_n\}$ of points of $S$ (a subset of $\mathbb{C}$ and $z_0\in S$) which converges to $z_0$ is verified that $f(z_n) \rightarrow f(z_0)$.
However,
$$\frac{\overline{z}}{z}=\frac{\frac{1}{n}}{\frac{1}{n}}=1 \qquad \text{and} \qquad \frac{\overline{z}}{z}=\frac{-\frac{i}{n}}{\frac{i}{n}}=-1.$$
Then, $f$ is not continuous in $z=0$. Remember that a function which verify the Cauchy-Riemann conditions can be not differentiable. It also must satisfied that $u$ and $v$ are differentiables in the point $(x_0,y_0)$ as functions defined on $S \subset \mathbb{R}^2$ with values in $\mathbb{R}$, where $f=u+iv$.