As mentioned in the related question, $\mathbb{R}^2$ is a too well-behaved space for such a function to exist.
If a topological space $X$ is locally path-connected and first countable, every function $f\colon X \to Y$, where $Y$ is any topological space, such that the composition $f\circ \alpha \colon I \to Y$ is continuous for all paths $\alpha \colon I \to X$ ($I \subset \mathbb{R}$ is some nondegenerate interval, open, half-open, or closed) is already continuous.
For if $f$ is discontinuous at $x_{\ast}\in X$, then by first countability there is a sequence $(x_n)_{n\in \mathbb{N}}$ in $X$ with $x_n \to x_{\ast}$ such that $f(x_n) \nrightarrow f(x_{\ast})$. By passing to a subsequence, we can assume that there is a neighbourhood $W$ of $f(x_{\ast})$ such that $f(x_n) \notin W$ for all $n$. Let $\{ V_k : k \in \mathbb{N}\}$ be a neighbourhood basis of $x_{\ast}$ consisting of path-connected sets, with $V_{k+1} \subset V_k$ for all $k$. By dropping some terms from the start of the sequence if necessary, we can assume that $x_n \in V_0$ for all $n$. We then define recursively $n_0 = 0$ and
$$n_{k+1} = \min \{ m > n_k : n \geqslant m \implies x_n \in V_{k+1}\}.$$
We can now replace $(x_n)$ by the subsequence $(x_{n_k})$, or equivalently assume that $n_k = k$ for all $k$.
Now we construct a path passing through the points $x_k$: Since $x_k$ and $x_{k+1}$ both lie in $V_k$, we can choose a path $\alpha_k \colon [2^{-k-1},2^{-k}] \to V_k$ with $\alpha_k(2^{-k-1}) = x_{k+1}$ and $\alpha_{k}(2^{-k}) = x_k$. Then define $\alpha \colon [0,1] \to X$ by
$$\alpha(t) = \begin{cases} x_{\ast} &, t = 0 \\ \alpha_k(t) &, 2^{-k-1} < t \leqslant 2^{-k}.\end{cases}$$
Since $\alpha_k(2^{-k-1}) = \alpha_{k+1}(2^{-k-1}) = x_{k+1}$, $\alpha$ is continuous on $(0,1]$, and since $\alpha([0,2^{-k}]) \subset V_k$, $\alpha$ is also continuous at $0$.
But $f(\alpha(2^{-k})) = f(x_k) \notin W$, so $f\circ \alpha$ is not continuous at $0$.
G1 continuous
geometrically continuous, tangents are collinear, may not have same magnitude
C1 continuous
parametric continuity ensured, notice that the magnitude of the vectors are equal
Note: I'm using 'Bezier' curves, so tangents will not overlap (2nd tangent is negative in Bezier), instead face in the opposite direction (as opposed to overlapping tangents in Hermite)
Best Answer
Consider the parameterized curve $\gamma(t)=(t^3,t^2)$. This parameterization is clearly $C^1$, because each component is (infinitely!) differentiable.
However, the graph of the curve looks like this:
As you can see, the left and right tangent vectors at the origin (where $t=0$) do not match, so the curve is not $G^1$.
This is possible because $\gamma'(0)=\left<0,0\right>$. That is, the parameterization comes gradually to a stop at $t=0$, which allows it to smoothly traverse a geometrically unsmooth curve.