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The Cuachy-Riemann equations tell us how the partial derivatives of $u$ and $v$ interact when $f = u + iv$ is holomorphic. A natural question is to ask whether these equations can guarantee that $f$ is holomorphic. The following example shows that the answer will not be an unqualified "yes".
Define $f : \C \to \C$ by \[f(z) = \begin{cases} 1 & xy = 0 \\ 0 & xy \ne 0 \end{cases}\] for all $z \in \C$. All of the partial derivatives \[(\partial_1 u)(0,0) \qquad (\partial_2 u)(0,0) \qquad (\partial_1 v)(0,0) \qquad (\partial_2 v)(0,0)\] exist and are equal to 0. However $f$ is not continuous at 0, because for instance \[\lim_{t \to 0} f(t + it) = \lim_{t \to 0} 0 = 0 \ne 1 = f(1)\] so it cannot be differentiable there.
We conclude from the above example that a function $f$ need not be differentiable at a point $z = x+iy$ just because the partial derivatives \[(\partial_1 u)(0,0) \qquad (\partial_2 u)(0,0) \qquad (\partial_1 v)(0,0) \qquad (\partial_2 v)(0,0)\] exist and satisfy the Cauchy-Riemann equations. What additional assumptions will guarantee $f$ is differentiable there? In our example, there is no open ball centered at zero inside which all the partial derivatives exist. A reasonable additional assumption would be to add the requirement that all the partial derivatives exist inside $\ball(z,r)$ for some $r > 0$. As the following example shows, this is not quite enough.
Define $f$ by \[f(z) = \begin{cases} (\overline{z})^2/z & z \ne 0 \\ 0 & z = 0 \end{cases}\] for all $z \in \C$. First of all, note that $f$ is not differentiable at $z = 0$ because \[\frac{f(z+0) - f(0)}{z} = \left( \dfrac{x-iy}{x+i y} \right)^2\] has different limits as $z \to 0$ along the real and imaginary axes.
For this function \[u(x,y) = \dfrac{x^3 - 3xy^2}{x^2 + y^2} \qquad v(x,y) = \frac{y^3 - 3x^2 y}{x^2 + y^2}\] whenever $(x,y) \ne 0$ and we calcalculate that the Cauchy-Riemann equations hold at $(0,0)$. However, for example, the partial derivative $\partial_1 u$ is not continuous at $(0,0)$. We have \[ (\partial_1 u)(x,y) = \begin{cases} \dfrac{x^4 - 3y^4 + 6x^2y^2}{(x^2 + y^2)^2} & (x,y) \ne (0,0) \\ 0 & (x,y) = (0,0) \end{cases}\] but $\lim\limits_{t \to 0} (\partial_1 u)(t,t) = 4$.
If, in addition to existing inside $\ball(z,r)$ for some $r > 0$ the partial derivatives are all continuous on $\ball(z,r)$ then we can conclude from the Cauchy-Riemann equations that $f$ is holomorphic at $z$.
Fix $z \in \C$ and $r > 0$. Suppose $u,v : \ball(z,r) \to \R$ satisfy all of the following properties.
Then $f(z) = u(x,y) + i v(x,y)$ is differentiable at $z$.
We already know that if $f'(z)$ exists then its derivative must be $(\partial_1 u)(x,y) + i (\partial_1 v)(x,y)$. Fix $\epsilon > 0$. We start by looking at $u(x+c,y+d)$ for $c^2 + d^2 < r^2$. Since $\partial_2 u$ exists at $(x+c,y)$ we can estimate that \[|u(x+c,y+d) - u(x+c,y) - d (\partial_2 u)(x+c,y) | < \epsilon |d|\] whenever $|d|$ is small enough. Similarly, since $\partial_1 u$ exists at $(x,y)$ we can estimate that \[|u(x+c,y) - u(x,y) - c (\partial_1 u)(x,y) | < \epsilon |c|\] whenever $|c|$ is small enough. Lastly, continuity of $\partial_2 u$ at $(x,y)$ guarantees \[| (\partial_2 u)(x+c,y) - (\partial_2 u)(x,y)| < \epsilon\] whenever $|c|$ is small enough. These three inequalities combined give \[|u(x+c,y+d) - u(x,y) - c (\partial_1 u)(x,y) - d (\partial_2 u)(x,y)| < (|c| + 2|d|)\epsilon\] whenever $|c|$ and $|d|$ are small enough. The same argument applied to $v$ gives \[|v(x+c,y+d) - v(x,y) - c (\partial_1 v)(x,y) - d (\partial_2 v)(x,y)| < (|c| + 2|d|)\epsilon\] so that \[\begin{aligned} & \; \bigg| \frac{u(x+c,y+d) + i v(x+c,y+d) - u(x,y) - i v(x,y)}{c+id} \\ & \qquad - \bigg( (\partial_1 u)(x,y) + i(\partial_1 v)(x,y) \bigg) \bigg| \\ \le & \; \frac{|u(x+c,y+d) - u(x,y) - c (\partial_1 u)(x,y) + d (\partial_1 v)(x,y)|}{\sqrt{c^2 + d^2}} \\ & \qquad + \frac{|v(x+c,y+d) - v(x,y) - d (\partial_1 u)(x,y) - c (\partial_1 v)(x,y)|}{\sqrt{c^2 + d^2}} \\ \le & \; 4 \frac{|c| + |d|}{\sqrt{c^2 + d^2}} \epsilon \end{aligned}\] if $|c|$ and $|d|$ are small enough, by the Cauchy-Riemann equations and the above estimates. Finally, note that \[\frac{|c| + |d|}{\sqrt{c^2 + d^2}} \le \frac{ 2 \max \{ |c|, |d| \} }{ \sqrt{2} \max \{ |c|, |d| \} } = \sqrt{2}\] giving the desired conclusion. $\square$