Partial derivatives do not imply differentiability

Counterexample

One could guess that a function of several variables is differentiable at a point when all its partial derivatives exist there. 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 $f(0, 0) = 0$. Both partial derivatives of $f$ exist at $(0, 0)$, and both are $0$. But $f$ is not continuous at $(0, 0)$, so $f$ is not differentiable at $(0, 0)$.

Remarks

The function is $0$ on both axes, so $f(h, 0) = f(0, h) = 0$ for each $h$, and both difference quotients at the origin are $0$. On the line $y = x$ the value is $1/2$, so $f$ is not continuous at the origin, as in the separately continuous function. A function that is differentiable at a point is continuous there. Even all directional derivatives are not enough. For $g(x, y) = x^2 y / (x^4 + y^2)$ off the origin and $g(0, 0) = 0$, each directional derivative at the origin exists, but $g = 1/2$ on the parabola $y = x^2$ minus the origin. If the partial derivatives exist near the point and are continuous at it, then the function is differentiable, by the criterion with continuous partial derivatives.

Explanations

L1 human

The partial derivatives look only along the two axes. A function can be flat on both axes and still do something wild between them, here on the diagonal. Differentiability asks for a good linear approximation in all directions at once, which is much more.