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The full question is as follows :

Let $g(x,y)$ be a harmonic function in a domain $D$ such that $g(x,y) >0$ for all $(x,y) \in D$. Consider the function $$h(x,y) = \dfrac{1}{g(x,y)}$$

Show that $$\dfrac{d^{2}h}{dx^{2}}+\dfrac{d^{2}h}{dy^{2}} = \dfrac{2}{[g(x,y)]^{3}}[(\dfrac{dg}{dx})^{2}+(\dfrac{dg}{dy})^{2}] ~~ ~\text{on} ~ D.$$

Let $h= \dfrac{1}{g}$, by the chain rule? We have \begin{align*} h_{xx}+h_{yy} &= (\dfrac{1}{g})_{xx}+(\dfrac{1}{g})_{yy}\\ &= (-\dfrac{1}{g^{2}}\cdot g_x)_x + (-\dfrac{1}{g^{2}}\cdot g_y)_y\\ &= -\dfrac{1}{g^{2}} \cdot g_{xx} +\dfrac{2}{g^{3}} \cdot g_x \cdot g_x -\dfrac{1}{g^{2}}g_{yy}+\dfrac{2}{g^{3}} \cdot g_y \cdot g_y\\ &= -\dfrac{1}{g^{2}} \cdot (g_{xx} + g_{yy}) + \dfrac{2}{g^{3}}(g_{x}^{2}+g_{y}^{2})\\ &= 0 + \dfrac{2}{g^{3}} \cdot (g_{x}^{2}+g_{y}^{2})\\ &= \dfrac{2}{g^{3}} \cdot (g_{x}^{2}+g_{y}^{2})\\ \end{align*}

I did the question by a way of motion, not knowing the reason behind how the chain rule works. I read on wiki and other links alike, but i do not know how chain rule on these kind of questions are applied. Can anyone show me how is the chain rule applied with intermediate steps or formulas to guide me? thanks!

So far i know chain rule is something like $\dfrac{d}{dx}[g(g(x))] = f^{'}(g(x)) \cdot g^{'}(x)$

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    Your analysis looks correct. What are you asking? – 2017-02-07
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    @Dr.MV I do not know how to use chain rule on $(\dfrac{1}{g})$ with respect to $x$. I merely did it by guess work and some assumption. – 2017-02-07
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    The expansion of $$\frac{d^2}{dx^2}\frac{1}{g}$$ is correct as is the one respect to $y$. – 2017-02-07
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    @Chinny84 Yeap i know it is correct, but i may know how to get to the answer explicitly? I do not know how to apply the formula for it – 2017-02-07

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To apply the chain rule to find the partial derivatives of $\frac1{g(x,y)}$ we introduce an new variables $u$ and $v$ such that $u=g(x,y)$ and $v(u)=\frac1u$..

Then, using the chain rule, we can write

$$\begin{align} \frac{\partial}{\partial x}\left(\frac{1}{g(x,y)}\right)&=\frac{\partial}{\partial x}\left(v(u(x,y))\right)\\\\ &=\color{blue}{\frac{\partial v(u)}{\partial u}}\color{red}{\frac{\partial u(x,y)}{\partial x}}\\\\ &=\color{blue}{\frac{\partial }{\partial u}\left(\frac1u\right)}\color{red}{\frac{\partial g(x,y)}{\partial x}}\\\\ &=\color{blue}{-\frac{1}{u^2}}\color{red}{\frac{\partial g(x,y)}{\partial x}}\\\\ &=\color{blue}{-\frac{1}{(g(x,y))^2}}\color{red}{\frac{\partial g(x,y)}{\partial x}} \end{align}$$

Can you proceed now?

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    Yes i can thank you so much for the clear explanation! – 2017-02-07
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    Did i manage to at your name on my comment @Dr.MV – 2017-02-07
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    You're welcome. My pleasure. -Mark – 2017-02-07
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    If I may add that you should use $v(g(x,y))$ rather than introducing two different functions. +1 in any case! – 2017-02-07
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    @Chinny84 Thank you. The reason for introducing $u$, is to write the chain rule without carrying a term $\frac{\partial }{\partial g(x,y)}$. – 2017-02-07
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    I am still not satisfied with the rationale - but the most important thing is the OP got exactly what your answer was pointing to! :) – 2017-02-07
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    @Chinny84 It is unconventional notation to write $\frac{\partial }{\partial g(x,y)}$. The introduction of a new variable $u=g(x,y)$ makes the analysis cleaner. ;-)) – 2017-02-07
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    I guess I can agree with the cleaner part, and I think you may be better off with $u$ rather than denoting $g = g(x,y)$ which may confuse things. thumbs up in any case.. – 2017-02-07
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    @Chinny84 Thanks Robert. I just sent you a LinkedIn invitation. -Mark – 2017-02-07