A demand function relates the quantity demanded of a good by a consumer with the price of the good. Thus we wish to find $Y = f(P_Y)$.
Setting up the optimization problem:
$$\max{U(X,Y)}$$
subject to: $$ I = P_x X + P_Y Y $$
where $I$ is income, $P_X$ is the price of good $X$, and $P_Y$ is the price of good $Y$.
Using the values you provided gives the optimization problem as:
$$ \max{ (XY + 10Y) } $$
subject to: $$ 100 = 1 \cdot X + P_Y Y $$
Setting this up as a Lagrange problem,
$$ L = XY + 10Y + \lambda (100 - X - P_Y Y )$$
Taking the first order conditions, we get:
$[X]:$ $\frac{ \partial U(X,Y) }{ \partial X} = Y - \lambda = 0$
$[Y]:$ $\frac{ \partial U(X,Y) }{ \partial Y} = X + 10 - \lambda P_Y = 0$
$[ \lambda ]:$ $\frac{ \partial U(X,Y) }{ \partial \lambda } = 100 - X - P_Y Y = 0$
Note, at this point you will usually take the second order conditions to ensure you have a maximum. Clearly you do have a maximum in this case since $U$ is strictly increasing in $X$ and $Y$.
Combining $[X]$ and $[Y]$ we get $X + 10 = Y P_Y$
We wish to get the demand for clothing, so we will solve for $X$ with the intention of substitution it into the budget constraint, $X = Y P_Y - 10$. Substituting into the constraint yields: $100 = 2 P_Y Y - 10$, or a final demand equation of:
$$ Y = \frac{45}{P_Y} $$
Finally, for a utility function to be quasi-linear, you must be able to express one utility as a linear function of one of the goods. Note in your case this may not be accomplished since you have an interaction between $X$ and $Y$. The reason quasi-linearity is nice is because it allows the expression of utility in terms of a numeraire good.
You don't need calculus for this, a simple figure suffices.
If I understand the problem correctly you want to maximize $U(x,y)$ under the constraints
$$x\geq0,\quad y\geq 0,\quad P_x\>x+P_y\>y\leq I\ .\tag{1}$$
The constraints $(1)$ define a triangle $T$ in the first quadrant as set of feasible points.
On the other hand $U(x,y)$ is a monotonically increasing function of $x+y$. Therefore we have to pick the vertex of $T$ where $x+y$ is maximal, and this is either the vertex $\bigl({I\over P_x},0\bigr)$ or the vertex $\bigl(0,{I\over P_y}
\bigr)$, depending on which of $P_x$ or $P_y$ is smaller.
Best Answer
I agree, it is hard to tell how to answer the question because I don't know what "competitive equilibrium" means. But, if I assume that it means that given prices, the consumption decision of each person is optimized and such that market clearing conditions hold, then you can solve using the following. Solve for the optimal $x_i, y_i$ in each of the following. Solve $$ \max_{x_a,y_a} y_a+ 50 \ln x_a \\ s.t. \; p_x x_a + p_y y_a \leq 200 p_x + 200 p_y $$ for person A, and $$ \max_{x_b,y_b} y_b + 150 \ln x_b \\ s.t. \; p_x x_b + p_y y_b \leq 100 p_x + 100 p_y $$ for person B. Use the method of Lagrange multipliers. The inequalities will hold with equality. Then, to pin down a solution, normalize the price of one good, say, by letting p_x = 1. Then apply the market clearing conditions (you'll only need to use one of them)
$$ x_a + x_b = 300 \\ y_a + y_b = 300.$$
(This gives the Walrasian (competitive) equilibrium in the general equilibrium framework.)