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Say we have a complex function $f(z)$. Then $z = x + iy$, so $z$ can be thought of as a function of $x$ and $y$. Thus we can take partial derivatives to get $$\frac{dz}{dx} = 1,\ \frac{dz}{dy} = i.$$

So by the chain rule, can we say that

$$\frac{df}{dz} = \frac{df}{dx}\cdot \frac{dx}{dz} + \frac{df}{dy}\cdot \frac{dy}{dz} = \frac{df}{dx} - i\cdot \frac{df}{dy}\ ?$$

However, I know that this goes against the Cauchy-Riemann equations since $\frac{df}{dz}$ is actually equal to $\frac{df}{dx}$ and $-i\cdot \frac{df}{dy}$. Am I missing something obvious here?

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    Are you sure the CR equations say that? The equations don't even mention $\frac {df}{dz}$ at all, they only talk about the partial derivatives of the real and imaginary parts of $f$. – 2017-01-25
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    http://mathworld.wolfram.com/Cauchy-RiemannEquations.html I believe it's used the derivation, but I don't understand where some of the derivatives here come from either. – 2017-01-25
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    @ThomasDoyle The derivatives $\frac{dz}{dx}$ and $\frac{dz}{dy}$ are the sandard $u_x, v_x$ for $f(z)=z$. In the second step, he seems to use the formula for the inverse function for these non-invertible functions. – 2017-01-25
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    The derivation in that link doesn't make much sense to me. – 2017-01-25

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The chain rule is requiring that the functions are differentiable, and $x,y$ are not differentiable with respect to $z$.

Also, another mistake in the computation is the fact that the formulas $$\frac{dx}{dz}=\frac{1}{\frac{dz}{dx}}$$ require $z$ to be a (locally) 1-1 function in $x$, and this is not happening.