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Let $k=\frac xy$ and $F(x,y)$ be a differentiable function. Now how to calculate $$\frac{\delta}{\delta k}\frac{F(x,y)}{y}$$ Will it be equal to $\frac{\delta F}{\delta x}?$

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    @Dr.SonnhardGraubner We don't use $\LaTeX$ here, we use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Very related, but not exactly the same thing. – 2017-01-29
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    @Dr.SonnhardGraubner Also, if you want to tell a new user to please conform to our formatting standards, the least you can do is link them to a resource, like I did. – 2017-01-29
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    done..have used mathematical symbols now – 2017-01-29

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I answered yesterday about a similar question. The fact is that you have to write your $F$ as a function of $k$, since you're deriving with respect to $k$.

In your case:

$$F(x, y) = F\left(ky, \frac{x}{k}\right)$$

Now you can take the derivative.

Let's make an example.

Be $F(x, y) = x^2 - 2y$ then your $k = \frac{x}{y}$ hence

$$x = ky ~~~~~~~ y = \frac{x}{k}$$

$$F(x, y) = x^2 - 2y \to (ky)^2 - 2\frac{x}{k} = k^2y^2 - \frac{2x}{k}$$

Now you can take the derivative with respect to $k$ because $k$ does figure directly in your function. Don't forget that you have a $y$ term in the denominator which has to be written as $x/k$ hence

$$\frac{d}{dk} \frac{F}{y} = \frac{d}{dk} \frac{k}{x}\left(k^2y^2 - \frac{2x}{k}\right)$$

$$\frac{d}{dk} \frac{F}{y} = \frac{d}{dk} \left(\frac{k^3 y^2}{x} - 2\right)$$

$$\frac{d}{dk} \frac{F}{y} = \frac{3k^2y^2}{x}$$

Now it's a call back to what $k$ is, and we find

$$\frac{3\frac{x^2}{y^2} y^2}{x} = 3x$$