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For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows:

$\frac {\partial f(x,y)} {\partial x} = \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$

$\frac {\partial f(x,y)} {\partial y} = \left(\frac {3 x} {4}\right)\left (\frac {k} {x^{(1/3)}}\right)^2 (2 y) \frac {1} {g(x,y) \sqrt {1 - (\frac {k y} {x^{(1/3)}})^2}}$

Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0) = x$.

I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$. Needless to say, I am not asking that they verify those suggestions, but in case someone would like, these are the inputs to Wolfram integrator:

1 / ( g(x,y as r) sqrt(1 - (k r / x^(1/3))^2) )

(3 t / 4) (k / t^(1/3))^2 (2 x) / ( g(x as t,y as x) sqrt(1 - (k x / t^(1/3))^2) )

Thanks in advance for your help.

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    @Alex: the other problem was that you registered *a new account* rather than registering your old account. I've merged the old accounts into your new registered one now. You should be able to finally keep track and comment on your own questions and answers.2011-02-16

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Hint: From your data one obtains $-{f_x\over f_y}={-2\over 3 k^2 y x^{1/3}}.$ It follows that the curves $f(x,y)=$const. satisfy the separable differential equation y'={-2\over 3 k^2 y x^{1/3}}.

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    Thank you very much, Professor. Your observation opened the way to the solution.2011-02-17