Answer
The curvature at the point $(1,1) $ is:
$$
\begin{aligned}
\kappa (1)&=\frac{12 }{\left(17\right)^{3 / 2}}
\end{aligned}
$$
Work Step by Step
We have the curve:
$$
y=x^{4},
$$
We first compute the ingredients needed
$$
y^{\prime}=4x^{3}, \ \ \ y^{\prime\prime}=12x^{2}
$$
So, the curvature formula of the curve is given by:
$$
\begin{aligned}
\kappa(x)&=\frac{\left|y^{\prime \prime}\right|}{\left[1+\left(y^{\prime}\right)^{2}\right]^{3 / 2}} \\
&=\frac{\left|12 x^{2}\right|}{\left(1+16 x^{6}\right)^{3 / 2}}
\end{aligned}
$$
Now, at the point $(1,1), \ x=1$, so, the curvature at the point $(1,1) $ is:
$$
\begin{aligned}
\kappa (1)&=\frac{\left|12 (1)^{2}\right|}{\left(1+16 (1)^{6}\right)^{3 / 2}}\\
&=\frac{12 }{\left(17\right)^{3 / 2}}
\end{aligned}
$$
