[tex]\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath[/tex]
The velocity at time [tex]t[/tex] is
[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath[/tex]
Take two vectors that point in the positive [tex]x[/tex] and positive [tex]y[/tex] directions, such as [tex]\vec\imath[/tex] and [tex]\vec\jmath[/tex]. The dot products of the velocity vector with [tex]\vec\imath[/tex] and [tex]\vec\jmath[/tex] are
[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta[/tex]
and
[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta[/tex]
We want the angles between these vectors to be 45º, for which we have [tex]\cos45^\circ=\frac1{\sqrt2}[/tex]. So
[tex]\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0[/tex]
[tex]\implies t(3ct-2b)=0[/tex]
[tex]\implies t=0\text{ or }t=\dfrac{2b}{3c}[/tex]
When [tex]t=0[/tex], the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for
[tex]\boxed{t=\dfrac{2b}{3c}}[/tex]
