Given a curve defined by $$\gamma(t)=(t-\frac{t^3}{3},t^2,t+\frac{t^3}{3})$$
Show that the $$κ=τ=\frac{1}{\left(1+t^{2}\right)^{2}}$$
Where $κ$ is the curvature, and $τ$ is the torsion of the curve.
By definition $$κ=\left|\frac{dT}{dt}\right|=\left|\frac{d}{dt}\frac{d\gamma(t)}{dt}\right|=\frac{1}{\sqrt{4\left(2t^{2}+1\right)}}=\frac{1}{2\sqrt{\left(2t^{2}+1\right)}}$$
So where was I wrong?
Problem with your calculations is that you are not considering the arc-length parametrization of your curve which those faulty definitions rely on, instead you are using a parametrically-defined space curve. You can use the formula on wikipedia whose derivation is hinted there: "For a parametrically-defined space curve in three dimensions given in Cartesian coordinates by $\gamma(t)=(x(t))$ $y(t), z(t),$ the curvature is" $$ \kappa=\frac{\sqrt{\left(z^{\prime \prime} y^{\prime}-y^{\prime \prime} z^{\prime}\right)^{2}+\left(x^{\prime \prime} z^{\prime}-z^{\prime \prime} x^{\prime}\right)^{2}+\left(y^{\prime \prime} x^{\prime}-x^{\prime \prime} y^{\prime}\right)^{2}}}{\left(x^{\prime 2}+y^{\prime 2}+z^{\prime 2}\right)^{\frac{3}{2}}} $$ resulting in exactly what you are looking for. I suggest reading the whole wikipedia page for a start.