[proofplan] We construct a rational function that behaves regularly along every straight line through the origin but has a nonzero limiting value along a curved path. The [directional derivative](/page/Directional%20Derivative) calculation reduces to a one-variable limit after writing an arbitrary direction as $v=(a,b)$. The discontinuity is detected by approaching the origin along the parabola $y=x^2$, where the function is constantly $\frac{1}{2}$ away from the origin. [/proofplan]

[step:Define the function and record its value at the origin] Define the function \begin{align*} f: \mathbb{R}^2 \to \mathbb{R} \end{align*} as follows. Set $f(0,0) := 0$. For every point $(x,y) \in \mathbb{R}^2 \setminus \{(0,0)\}$, set \begin{align*} f(x,y) := \frac{x^2 y}{x^4 + y^2}. \end{align*} The denominator is positive whenever $(x,y) \neq (0,0)$, since $x^4 \geq 0$, $y^2 \geq 0$, and $x^4+y^2=0$ would force $x=0$ and $y=0$. Hence $f$ is well-defined on all of $\mathbb{R}^2$. [/step]

[step:Compute every directional derivative at the origin]Let $v=(a,b) \in \mathbb{R}^2$ be arbitrary, where $a,b \in \mathbb{R}$. For $t \in \mathbb{R} \setminus \{0\}$, the directional difference quotient is \begin{align*} \frac{f((0,0)+t(a,b))-f(0,0)}{t} = \frac{f(ta,tb)}{t}. \end{align*} If $b \neq 0$, then for all sufficiently small $t \neq 0$ we have $(ta,tb) \neq (0,0)$, and therefore \begin{align*} \frac{f(ta,tb)}{t} = \frac{1}{t}\frac{(ta)^2(tb)}{(ta)^4+(tb)^2}. \end{align*} Simplifying the powers of $t$ gives \begin{align*} \frac{f(ta,tb)}{t} = \frac{a^2 b}{t^2 a^4 + b^2}. \end{align*} Since $b^2>0$, the denominator tends to $b^2$, so \begin{align*} D_v f(0,0) = \lim_{t \to 0} \frac{f(ta,tb)}{t} = \frac{a^2}{b}. \end{align*} If $b=0$, then $v=(a,0)$. For every $t \neq 0$, if $ta \neq 0$ then $f(ta,0)=0$ by the defining formula, and if $ta=0$ then $(ta,0)=(0,0)$ and $f(ta,0)=f(0,0)=0$. Hence \begin{align*} \frac{f(ta,0)}{t} = 0 \end{align*} for every $t \neq 0$, and consequently $D_v f(0,0)=0$. This also covers the zero direction $v=(0,0)$. Thus $D_v f(0,0)$ exists for every $v \in \mathbb{R}^2$.[/step]

[guided]We must check the directional derivative in an arbitrary direction, so fix a vector $v=(a,b) \in \mathbb{R}^2$, where $a,b \in \mathbb{R}$. By definition, the directional derivative at the origin is the limit of the one-variable quotient \begin{align*} \frac{f((0,0)+t(a,b))-f(0,0)}{t} = \frac{f(ta,tb)}{t} \end{align*} as $t \to 0$ through nonzero [real numbers](/page/Real%20Numbers). The calculation splits according to whether the line has nonzero $y$-component. First suppose $b \neq 0$. Then $(ta,tb) \neq (0,0)$ for every $t \neq 0$, so the rational formula for $f$ applies: \begin{align*} \frac{f(ta,tb)}{t} = \frac{1}{t}\frac{(ta)^2(tb)}{(ta)^4+(tb)^2}. \end{align*} Now we simplify carefully. The numerator inside $f$ is $t^3 a^2 b$, while the denominator inside $f$ is $t^4 a^4+t^2 b^2$. Dividing by the outside factor $t$ gives \begin{align*} \frac{f(ta,tb)}{t} = \frac{a^2 b}{t^2 a^4+b^2}. \end{align*} Because $b \neq 0$, the denominator tends to $b^2$, not to $0$. Therefore the quotient has the finite limit \begin{align*} \lim_{t \to 0}\frac{f(ta,tb)}{t} = \frac{a^2 b}{b^2} = \frac{a^2}{b}. \end{align*} Next suppose $b=0$. Then the direction is horizontal: $v=(a,0)$. For every $t \neq 0$, the point on the line is $(ta,0)$. If $ta \neq 0$, the defining formula gives \begin{align*} f(ta,0) = \frac{(ta)^2 \cdot 0}{(ta)^4+0^2} = 0. \end{align*} If $ta=0$, then the point is exactly $(0,0)$, and $f(0,0)=0$ by definition. Thus in every case \begin{align*} \frac{f(ta,0)}{t} = 0 \end{align*} for all $t \neq 0$. Hence the limit exists and equals $0$. This includes the zero vector $v=(0,0)$, for which the quotient is also identically $0$. Since the vector $v=(a,b)$ was arbitrary, every directional derivative $D_v f(0,0)$ exists.[/guided]

[step:Approach the origin along a parabola to disprove continuity] Define the path \begin{align*} \gamma: \mathbb{R} \to \mathbb{R}^2, \qquad \gamma(s) := (s,s^2). \end{align*} For every $s \in \mathbb{R} \setminus \{0\}$, we have $\gamma(s) \neq (0,0)$, so \begin{align*} f(\gamma(s)) = f(s,s^2) = \frac{s^2 s^2}{s^4+(s^2)^2}. \end{align*} Since $(s^2)^2=s^4$, this simplifies to \begin{align*} f(\gamma(s)) = \frac{s^4}{2s^4} = \frac{1}{2}. \end{align*} Also $\gamma(s) \to (0,0)$ as $s \to 0$, because \begin{align*} |\gamma(s)-(0,0)| = |(s,s^2)| = \sqrt{s^2+s^4} \to 0. \end{align*} If $f$ were continuous at $(0,0)$, then $f(\gamma(s))$ would converge to $f(0,0)=0$ as $s \to 0$. Instead $f(\gamma(s))=\frac{1}{2}$ for every $s \neq 0$. Therefore $f$ is not continuous at $(0,0)$. [/step]

[step:Conclude the example has the required properties] The function $f: \mathbb{R}^2 \to \mathbb{R}$ constructed above has a finite directional derivative $D_v f(0,0)$ for every $v \in \mathbb{R}^2$, but it is not continuous at $(0,0)$. This proves the theorem. [/step]