I am working on the question below.
$f(x,y)= \sqrt{5x^2+5y^2} $
First find $f_x$,$f_{xy}$ and $f_y$,$f_{yx}$ (see below)
But the last part of the question asks if this satisfies Laplaces Equation $f_{xx}+f_{yy}=0$
My question is : Is there an easier way to answer this question knowing the results of the first part of the question. My reason being that $f_{xx}$ and $f_{yy}$ seem to be too cumbersome.
$f_x(x,y) =\frac{5x}{\sqrt{5x^2+5y^2}} $
$f_{xy}(x,y) =\frac{25xy}{(\sqrt{5x^2+5y^2})^{-\frac{3}{2}}} $
$f_y(x,y) =\frac{5y}{\sqrt{5x^2+5y^2}} $
$f_{yx}(x,y) =\frac{25xy}{(\sqrt{5x^2+5y^2})^{-\frac{3}{2}}} $
I doubt you are supposed to use it, but the maximum principle works. Roughly it says that a harmonic function cannot have a true local maximum/minimum.
As you can tell, this function is nonnegative and has the value $0$ at $(0,0)$, so it does have a local minimum. Moreover it is not constant (e.g. since its value is $\sqrt5$ at $(1,0)$), thus by the maximum principle, it cannot be harmonic.
You can also use the mean value property, which says that if $f$ is harmonic on some open set containing the unit disk, $B(0,1)$, then for some appropriate positive constant $c$,
$$f(0,0) = c\cdot \int_{B(0,1)}f\,dV.$$ Combine this with the fact that $f(0,0) = 0$ and that $f$ is positive almost everywhere, you get that the lhs is zero while the rhs is positive, which is impossible so $f$ cannot be harmonic.
If you are interested, you can read more about harmonic functions on wikipedia.