<s>[INST] Maximum and minimum function on area Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $. 
I will grateful for hints and yours help.
 [/INST] What could be said about this "skewed superellipse"  $ \ 3x^4 \ + \ xy \ + \ y^4 \ = \ 6 \ $ is that since the curve has symmetry about the origin [if a point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x,-y) \ $ ], we should expect that extrema are located at corresponding points in opposing quadrants.  user64494  finds, by implicit differentiation of the curve equation,  the slope of a tangent line to be  $ \ \frac {12x^3+y}{4y^3+x} \ $ , and from the linear function to be extremized, the "useful" equation
$$\frac {12x^3 \ + \ y} {4y^3 \ + \ x} \ \ = \ \ \frac{3}{14} \ \ . $$
As the OP, Thomas, indicated that this was an exam problem, I wondered whether some indication was given that the student was to locate a rational point on this curve in answering the question, since the time would be severely (hopelessly?) limited to find real zeroes of the related polynomial without a computational aid.  If it were given that the coordinates of a rational point have the same denominator, so that $ \ x \ = \frac{a}{c} \ \ , \ \ y \ = \frac{b}{c} \ \ , $ we would be able to write
$$\frac {12 \left( \frac{a}{c} \right)^3 \ + \ \frac{b}{c} } {4\left( \frac{b}{c} \right)^3 \ + \frac{a}{c} } \ \ = \ \ \frac {12 a^3 \ + \ bc^2 } {4 b^3 \ + \ ac^2 } \ \ = \ \ \frac{3}{14} \ \ . $$
We might then begin a search with $ \ c  \ = \ 2 \ $ (much as we do in "hunting" rational zeroes of a polynomial), giving us
$$ \  \frac {12 a^3 \ + \ 4b  } {4 b^3 \ + \ 4a } \ \ = \ \ \frac{3}{14} \ \ . $$
Since we need $ \ a \ $ and $ \ b \ $ to be integers, we may require that $ \ 6 a^3 \ + \ 2b \ \ = \ \ 3k \ $ and $ \ 2 b^3 \ + \ 2a \ \ = \ \ 14k \ \ , $ with $ \ k \ $ being a positive integer.  It takes only a little effort to find $ \ 6  · 1^3 \ + \ 2·3 \ \ = \ \ 12 \ \ = \ \ 3·4 \ $ ; we then note that  $ \ 14 · 4 \ \ = \ \ 56 \ \ = \ \ 2 ·3^3 \ + \ 2·1 \ \  \ \ . $
This gives us the desired ratio $ \ \frac{12}{56} \ \ = \ \ \frac{3}{14} \ \ . $  As all of the exponents in the slope expression are odd, the signs of both $ \ a \ $ and $ \ b \ $ can both be changed without altering the sign of the ratio.  This then yields two rational points $ \ \left( \ \frac{1}{2} \ , \ \frac{3}{2} \ \right) $ and $ \ \left( \ -\frac{1}{2} \ , \ -\frac{3}{2} \ \right) \ \ , $  as expected from the symmetry of the curve.  The maximum and minimum of $ \ f(x,y) \  = \ 3x \ + \ 14y \ $ occur, respectively, for those two points as $ \ \frac{45}{2} \ $ and $ \ -\frac{45}{2} \ $ .  A representative graph is shown below.  A calculation of this sort may have been possible, with the clues mentioned, in the available time.

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I can't help feeling that this is one of those exam problems, if it is correctly reproduced here, that the problem-poser had not worked out in advance to estimate how much work would be necessary.  (I've had to reject many prospective "interesting" exam problems after seeing the amount of solution-writing that would be called for...)  An optimization problem for an exam would be rather more tractable if the curve were a "rotated ellipse", say,   $ \ 3x^2 \ + \ xy \ + \ y^2 \ = \ C \ $ .
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