kinetic description of an autocatalytic system
Let's first assign rate constants to each reaction for completeness:

$\ce{2X + Y &lt;=&gt;[k_1][k_{-1}] 3X}$

$\ce{A -&gt;[k_2] Y}$

$\ce{X-&gt;[k_3] B}$


Then let's consider what's happening overall. A is converting to Y, Y is converting to X, and X is converting to B:
$\ce{A -&gt;[k_2] Y &lt;=&gt;[k_1X^2][k_{-1}X^2] X -&gt;[k_3] B}$
Y and X may be in equilibrium, and X catalyzes the X-Y interconversion.
Since X and Y are intermediates, I'm guessing that you are writing the rate equations for them in order to do a steady-state approximation?
Regardless of the purpose, we can see that X is created only in the reaction $\ce{Y -&gt; X}$, which has $\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]$. However, it is consumed in two possible reactions, the back reaction $\ce{X -&gt; Y}$ and the forward reaction $\ce{X-&gt; B}$. So the consumption rate is $-\frac{d[\ce{X}]}{dt}=k_{-1}[\ce{X}]^3+k_3[\ce{X}]$. Combining the two equations, we find that
$$\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]-k_{-1}[\ce{X}]^3-k_3[\ce{X}].$$
This equation is similar to yours, but complete with rate constants and including the extra term for the reverse reaction. If that reaction is negligible, that term would drop out and give an equation like yours.
For Y, you can use the same process to again get a slightly longer equation than what you have written, because it includes an extra term for formation of Y from X in the back reaction. Clearly the derivation of the equations you provided assumes that only the forward reactions are significant.