Unconstrained optimization - parameters' values for global maximum Consider the following function
$f(x,y)=p \cdot x^a \cdot y^b - w_xx-w_yy$
with $(p,x,y,w_x,w_y) \in \mathbb{R^5_+} $ (this is because $f(x,y)$ is a profit function)
(a) Solve the first-order conditions in order to find $(x,y)$ that maximizes $f(x,y)$.
Here's my answer (it's a bit long, so I'll skip some steps).
$$\begin{cases} f_x = 0 \therefore p \cdot a \cdot x^{a-1} \cdot y^b = w_x \,\, \dots I \\ f_y = 0 \therefore p \cdot b \cdot x^a \cdot y^{b-1} = w_y \,\, \dots II\end{cases}$$
Since $x \neq 0, y \neq 0$, we can divide the equations and get:
$$ \frac{a}{b} \cdot x^{-1} \cdot y = \frac{w_x}{w_y} \therefore y = \frac{b}{a} \cdot \frac{w_x}{w_y} \cdot x \,\, \dots III$$
Substituting back in equation (I), we find:
$$ \large{x^* = \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1}}$$
and
$$ \large{y^* = \frac{b}{a} \cdot \frac{w_1}{w_2} \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} }$$
(b) Use the second-order conditions to determine the parameters' $a,b,p,w_x,w_y$ values such that the solution for the optimization problem is a global maximum.
The Hessian will be:
$$ H_{f(x,y)}= \begin{bmatrix} p \cdot a \cdot (a-1) \cdot x^{a-2} \cdot y^b & p \cdot a \cdot b \cdot x^{a-1} \cdot y^{b-1} \\ p \cdot a \cdot b \cdot x^{a-1} \cdot y^{b-1} & p \cdot b \cdot (b-1) \cdot x^a \cdot y^{b-2}\end{bmatrix} $$
In order for $f(x,y)$ to be strictly concave, $|H_2| < 0, |H| > 0$ where $H_2$ is the sub-matrix of $H$ that is found by removing the second column and second row.
For $|H|$, we have:
$$ |H| = p^2 \cdot a \cdot (a-1) \cdot b \cdot (b-1) \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 - p^2\cdot a^2 \cdot b^2 \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 \therefore \\\\ |H| = p^2 \cdot a \cdot b \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 \cdot [ (a-1) \cdot (b-1)-ab ]$$
Now if we make $p^2 \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 = k > 0$, all we have to worry about is:
$a \cdot b \cdot [(a-1)(b-1) -ab] = ab \cdot [ab - a -b + 1 - ab] = ab \cdot (1-a-b)$
So our first inequality is $ab \cdot (1-a-b) > 0$.
Now, substituting our optimal values into $|H_2|$ we have:
$|H_2(x^*,y^*)| = p \cdot a \cdot (a-1) \cdot \left( \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} \right)^{a-2} \cdot \left( \frac{b}{a} \cdot \frac{w_1}{w_2} \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} \right)^b \\\\ \therefore |H_2(x^*,y^*)| = p \cdot a \cdot (a-1) \cdot \left( \frac{b}{a} \cdot \frac{w_x}{w_y} \right)^b \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_y^b \cdot w_x^{1-b}\right)^{\frac{a+b-2}{a+b-1}} $
Now since $p > 0, w_x > 0, w_y > 0$, I'll remove them from the equation to declutter it a bit and I'll also drop the $(x^*,y^*)$. 
$$ \large{|H_2| = a \cdot (a-1) \cdot b^{-b} \cdot a^{-b} \cdot a^{\frac{(b-1)(a+b-2)}{a+b-1}} \cdot b^{\frac{b \cdot (a+b-2)}{a+b-1}} \therefore \\ |H_2| = (a-1) \cdot a^{\frac{(1-b)(a+b-1) + (b-1)(a+b-2)}{a+b-1}} \cdot b^{\frac{-b(a+b-1) + b(a+b+2)}{a+b-1}} \therefore \\ |H_2| = (a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{3b}{a+b-1}} \therefore \\\\ |H_2| = (a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{b}{a+b-1}} \cdot \left( b^{\frac{b}{a+b-1}} \right)^2}$$
Now, because the last term is always positive, all we need is:
$$ \large{(a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{b}{a+b-1}} < 0 } $$
Without imposing further restrictions I have no idea how to continue. If we impose $a>0,b>0$, than our problem can be reduced to
$$ \begin{cases} -a-b+1 >0 \\ (a-1) \cdot a < 0 \end{cases} $$
and our final solution is $$ \boxed{\boxed{0 < a < 1, \\ 0 < b < 1-a, \\(p,w_x,w_y) \in \mathbb{R_+^3} }} $$
Is my solution correct? Did I miss something that would make the calculations simpler or at least feasible?
Thanks in advance!

For the second part, you only need the hessian to be negative semidefinite everywhere and strictly negative definite at the candidate point found from the first part. This means that you have a concave function with only one local maximum, that is, this local minimum is unique in a small neighborhood. Thus, it is the unique global maximum of the function due to its concavity.
Also, the determinant of the symmetric matrix $H$ must be nonpositive as a necessary condition for negative semidefiniteness. 
