Separately continuous but not continuous function

Counterexample

One could guess the following. A function of two variables is continuous when it is continuous in each variable for each fixed value of the other variable. This is false. Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be given by $f(x, y) = \dfrac{xy}{x^2 + y^2}$ for $(x, y) \ne (0, 0)$ and $f(0, 0) = 0$. For each fixed $b$, the function $x \mapsto f(x, b)$ is continuous on $\mathbb{R}$, and for each fixed $a$, the function $y \mapsto f(a, y)$ is continuous on $\mathbb{R}$. But $f$ is not continuous at $(0, 0)$.

Remarks

For $b \ne 0$, the function $x \mapsto f(x, b)$ is a quotient of polynomials with a denominator that is never $0$. For $b = 0$, it is the constant $0$. The same holds for the other variable. On the line $y = x$ with $x \ne 0$, the value is $f(x, x) = 1/2$. So $f(x, x) \to 1/2 \ne f(0, 0)$ as $x \to 0$, by the test with different paths.

Explanations

L1 human

Separate continuity checks only the two directions along the axes. On both axes this function is $0$, so nothing goes wrong there. On the diagonal it is $1/2$ everywhere except at the origin, so the function jumps there.