Question: How do we calculate
$$2{\vec d\cdot \vec m\over \vec m\cdot \vec m}\vec m-\vec d$$

given the data ${\vec d = (1, 0)}$ and ${\vec m = (1, 1)}$?

The answer should be $(0, 1)$, but I don't know why.

Answer: 
If you can only use L'Hospital, rewrite $$\frac{(\cos{x})^2}{\sin{(x^2)}}-\frac{1}{x^2}=\frac{x^2(\cos{x})^2-\sin{(x^2)}}{x^2\sin{(x^2)}}=\frac uv$$ Now, you will have a funny time since you must apply the rule four times.

I give below the successive derivatives of $v=x^2\sin{(x^2)}$
$$v'=2 x \sin \left(x^2\right)+2 x^3 \cos \left(x^2\right)$$
$$v''=2 \sin \left(x^2\right)+10 x^2 \cos \left(x^2\right)-4 x^4 \sin \left(x^2\right)$$
$$v'''=24 x \cos \left(x^2\right)-8 x^5 \cos \left(x^2\right)-36 x^3 \sin \left(x^2\right)$$
$$v''''=-156 x^2 \sin \left(x^2\right)+\color{red}{24 \cos \left(x^2\right)}+16 x^6 \sin
 \left(x^2\right)-112 x^4 \cos \left(x^2\right)$$ where, for first time, you see a term which is not going to $0$.
You can simplify your life by noting that 

$$\frac{1}{\sin x}-\frac{1}{x}=x\frac{x}{\sin x}\frac{x-\sin x}{x^3}\to 0$$

So $$\frac{(\cos x)^2}{\sin (x^2)}-\frac{(\cos x)^2}{x^2}\to 0$$

and the limit becomes 
$$\frac{(\cos x)^2}{x^2}-\frac{1}{x^2}$$
 which is much easier, even if you use the hospital rule.