Here is the question:
Find $a$ and $b$, letting $$f(x)=\sin x-\frac{x+ax^3}{1+bx^2}$$ be the infinitesimal of the highest order when $x \to 0$, and find that order.
According to the key, $a=-\frac{7}{60}$, $b=\frac{1}{20}$, and the highest order can be reached is $7$.
I have used the Maclaurin expansion of $\sin x$, but I cannot understand how can things like $-\frac{x^3}{3!}$ and $\frac{x^5}{5!}$ be cancelled. It seems what we have is just $\frac{x+ax^3}{1+bx^2}$ and its highest order is only 1.
The Maclaurin expansion of $\sin x$ is, as you know, $x - x^3/6 + x^5/120 - x^7/5040 + \cdots$.
You want to choose $a$ and $b$ so that the Maclaurin expansion of $(x+ax^3)/(1+bx^2)$ matches this for as many terms as possible.
You wouldn't want to find that expansion by differentiating repeatedly, but you don't have to - rather you can just write out the denominator as
$$ {1 \over 1+bx^2} = 1 - bx^2 + b^2 x^4 - b^3 x^6 + \cdots $$
and multiply out:
$$ \left( x + ax^3 \right) \left( 1 - bx^2 + b^2 x^4 + \cdots \right) = x + (a-b) x^3 + (b^2 - ab) x^5 + \cdots $$
so now you want to find $a, b$ such that $a-b = -1/6, b^2-ab = 1/120$. These turn out to be the values of $a$ and $b$ that are given in the key.
But if you try to do this matching the $x^7$ terms as well, you'll find that the resulting system of three equations in two unknowns has no solution. So this is the best you can do.