Deriving demand functions given utility
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.