How to determine the isotherms of given temperature field?

Use the definition of "isotherm". For a given temperature, C, you are points where the temperature is C: [itex]x^2- y^2+ 8y= C[/itex]. Now, if you are asking "how can I write that as a function I know how to graph quickly?", the answer is, as so often for things like this, complete the square.

8/2= 4 and [itex]4^2= 16[/tex] so [tex]y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16[/tex]

We can write [itex]x^2- y^2+ 8y= x^2- (y^2+ 8y)= x^2- ((y- 4)^2- 16)= x^2- (y- 4)^2- 16[/itex] so that [itex]x^2- y^2+ 8y= C[/itex] becomes [itex]x^2- (y- 4)^2- 16= C[/itex] or [itex]x^2- (y- 4)^2= C+ 16[/itex]. To get that to "standard form", divide through by C+ 16:
[tex]\frac{x^2}{C+ 16}- \frac{(y- 4)^2}{C+ 16}= 1[/tex]

That is an equilateral hyperbola with center at (0, 4), vertices at [tex](\sqrt{C+ 16}, 4)[/tex] and [tex](-\sqrt{C+ 16}, 4), and asymptotes y= x+ 4 and y= -x+ 4.

 

Not when there is NO such function, as here.