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
Hint: You solved the equation incorrectly.
$$\frac{1}{(10-2y)} = \frac{P_x}{P_y}$$
Flipping numerators and denominators.
$$10-2y = \frac{P_y}{P_x}$$
$$10-\frac{P_y}{P_x} = 2y$$
$$y=5-\frac{P_y}{2P_x}$$
Then $$x=\frac{I-P_y\cdot y}{P_x}=\frac{I}{P_x}-\frac{P_y}{P_x}\cdot y$$
Inserting the expression of $y$
$$x=\frac{I}{P_x}-\frac{P_y}{P_x}\cdot \left(5-\frac{P_y}{2P_x}\right)$$
$$x=\frac{I}{P_x}-\frac{5P_y}{P_x}+\frac{P_y^2}{2P_x^2}$$