Why is the derivative of f equal to the sum of its partials along its components?

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I was trying to understand the development of the solution in this answer, where $\overline{f(z)}f'(z)dz = (u - iv)(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial y})(dx + idy)$. The first term is pretty obvious, as it is just the conjugate. However, where does the equality

$$f'(z)dz = \left(\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}\right)(dx + idy)$$ come from?

I know that we can write the derivative of an analytic function $f$ as

$$f'(z) = \frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y} = \frac{\partial v}{\partial y} + i \frac{\partial v}{\partial x}$$

but the above term does not seem to follow algebraically. Where does this term come from?

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This comes from the Cauchy-Riemann equations, since $f$ is analytic:

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

so it follows from my second expression.