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.