Your equation is
$$\tag{1}
(1 + x^3)y^2 + 4 \int_{2x}^{xy}(5x^2+t^2 )^{0.5} dt = 112.
$$
Note that if $y=2$, then the integral in $(1)$ is zero since the lower and upper limits of integration are the same. You can then solve for $x$ and discover that it is 3 when $y=2$.
Of course, you'll want to implicitly differentiate equation $(1)$. But there is a subtlety here when differentiating the integral. The integrand in the term $$\Phi(x)=\int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt$$
is a function of $x$ and $t$; you cannot use the Fundamental Theorem of Calculus (which requires that the integrand is a function of $t$ only) directly to find its derivative.
In particular, even if you split the integral into two parts
$$
\int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt=
\int_{2x}^{0} (5x^2+t^2)^{1/2}\,dt+
\int_{0}^{xy} (5x^2+t^2)^{1/2}\,dt,
$$
you cannot say, for example, that
$$
{d\over dx} \int_{0}^{xy} (5x^2+t^2)^{1/2}\,dt=
(5x^2+(xy)^2)^{1/2}\cdot {d\over dx}(xy).
$$
However, to find the derivative of $\Phi$, you can use the technique of differentiation under the integral sign. Using this rule gives:
$$\eqalign{
{d\over dx} \Phi(x)&=
{d\over dx} \int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt\cr&=
(5x^2+(xy)^2)^{1/2} (y+xy')-2(5x^2+4x^2)^{1/2}\,
+\int_{2x}^{xy}{\partial\over\partial x} (5x^2+t^2)^{1/2}\,dt
}
$$
(note the rule gives what you would obtain if you just applied FTOC plus an integral term).
After implicitly differentiating both sides of $(1)$,
you should wind up with
$$\tag{2}3x^2y^2+(1+x^3)2y\cdot y'+4\textstyle {d\over dx} \Phi(x) .
$$
When you evaluate $(2)$ at $y=2$, $x=3$, the integral term is zero, and you can then solve for $y'|_{y=2}$; thus, obtaining the posted solution.