Relation between $f(x,y)_{yy}$ and $f(x,-y)_{yy}$ The exercise I am talking about is exercise 6 of section 2.1.2 of Ahlford's Complex Analysis. It needs to be proven that if $f(z)$ is harmonic, then $f(\bar{z})$ is also harmonic. What I tried was to test if the second function is also harmonic based on the obtained conclusions of the first one being harmonic (also, just changing the sign of one of the values that go to a function doesn't seem to be something that would make the function non-harmonic).
My main problem appears when trying to obtain which is the second derivative over y of the second function $\frac{d^2 f(x,-y)}{dy²}$. I tried to do it by considering that the term $-y$ is just a function of $y$, let's say $g(y) = -y$, and so by using the chain rule we obtain:
$$ \frac{d f(x,g(y))}{dy} = \frac{df}{dg} \cdot \frac{dg}{dy} = -\frac{df}{dg}$$
From this, I don't know how can the last term be obtained based on the previous calculations (this is, the derivative over the new declared function).
 A: Let's put this in context by reviewing some single variable calculus: If we have $h(x)$, then $\dfrac{\mathrm d}{\mathrm dx}h(-x)=-h'(-x)$ by the chain rule. If you like, we can write this in terms of your $g$ as $\dfrac{\mathrm d}{\mathrm dx}h(g(x))=g'(x)h'(g(x))$.
Now, if we have a function of multiple variables like $f(x,y)$, the partial derivative (note I use $\partial$ rather than $\mathrm d$) $\dfrac{\partial}{\partial y}$ treats $x$ as a constant. So define $h_x(y)=f(x,y)$ and then you can apply the single variable chain rule discussed above to get that the desired quantity is $\dfrac{\mathrm d}{\mathrm dy}h_x(-y)=-h_x'(-y)$.
The tricky part of the notation here is that we would probably write that desired quantity as $\dfrac{\partial}{\partial y}f(x,-y)$, but this is kind of ambiguous: do we mean to substitute in $-y$ and then take the partial derivative of the resulting expression, or to take the partial derivative of $f$ and evaluate it at a point $(x,-y)$? One workaround would be to use a different notation when we want to evaluate partials of $f$ as points, like $f_2(a,b)=\left.\dfrac{\partial}{\partial y}f(x,y)\right|_{(x,y)=(a,b)}$. Then we have $\dfrac{\partial}{\partial y}f(x,-y)=\dfrac{\mathrm d}{\mathrm dy}h_x(-y)=-h_x'(-y)=-f_2(x,-y)$.
However, I want to emphasize that the detour through $h$ above was just to make it clear what it is we're calculating; it's not necessary. The answer to your question about the meaning of $\dfrac{\partial f}{\partial g}$ is that, in context (because you had put $g$ into the second input slot of $f$), it means what I'm calling $f_2(x,g(y))$, so that your chain rule answer of $-\dfrac{\partial f}{\partial g}$ is correct and agrees with my $-f_2(x,-y)$.
