Use the chain rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\cdot\dfrac{\mathrm dt}{\mathrm dx}[/tex]
So we have
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
[tex]x=t^2+4\implies\dfrac{\mathrm dx}{\mathrm dt}=2t[/tex]
[tex]y=t^3+3t\implies\dfrac{\mathrm dy}{\mathrm dt}=3t^2+3[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3t^2+3}{2t}[/tex]
Now write [tex]f(t)=\dfrac{\mathrm dy}{\mathrm dx}[/tex]. Then by the chain rule,
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{\mathrm dy}{\mathrm dx}\right]=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\cdot\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
so that
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac{\mathrm d}{\mathrm dt}\left[\frac{3t^2+3}{2t}\right]}{2t}=\dfrac{3(t^2-1)}{4t^3}[/tex]
The curve is concave upward when the second derivative is positive:
[tex]\dfrac{3(t^2-1)}{4t^3}>0\implies t^2>1\implies\sqrt{t^2}>\sqrt1\implies|t|>1[/tex]
or equivalently, when [tex]t<-1[/tex] or [tex]t>1[/tex].
