The partial derivatives $f_x$ and $f_y$ are themselves functions of $x$ and $y$, so you can differentiate them again. That gives four second partials: $f_{xx}$, $f_{yy}$ and two mixed partials, $f_{xy}$ and $f_{yx}$.
Key idea
$f_{xy} = (f_x)_y$: differentiate in $x$ first, then $y$. In Leibniz notation this is $\frac{\partial^2 f}{\partial y\,\partial x}$ (read right to left). Clairaut's theorem: if $f_{xy}$ and $f_{yx}$ are continuous near a point, they are equal there.
Why it works
Both mixed partials come from the same "double difference"
$$\Delta = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b).$$
Group it as $$[f(a + h, b + k) - f(a, b + k)] - [f(a + h, b) - f(a, b)]$$ and it measures how an $x$-change depends on $y$; group it the other way and it measures how a $y$-change depends on $x$. Dividing by $hk$ and letting $h, k \to 0$ in one order gives $f_{xy}$, in the other order $f_{yx}$. When the second partials are continuous, the order of the limits does not matter, so the two agree.
Meaning: $f_{xx}$ is the concavity of the trace in the $x$-direction; $f_{xy}$ measures twist: how the $x$-slope changes as you step in $y$.
Meaning: $f_{xx}$ is the concavity of the trace in the $x$-direction; $f_{xy}$ measures twist: how the $x$-slope changes as you step in $y$.
Worked examples
Worked example: All four second partials
Find the second partials of $f(x, y) = x^3y^2 - 4xy^3$.
- $f_x = 3x^2y^2 - 4y^3$ and $f_y = 2x^3y - 12xy^2$.
- $f_{xx} = 6xy^2$ and $f_{yy} = 2x^3 - 24xy$.
- $f_{xy} = (f_x)_y = 6x^2y - 12y^2$ and $f_{yx} = (f_y)_x = 6x^2y - 12y^2$: equal, as Clairaut promises.
Answer: $f_{xx} = 6xy^2$, $f_{yy} = 2x^3 - 24xy$, $f_{xy} = f_{yx} = 6x^2y - 12y^2$
Worked example: With an exponential
Find $f_{xx}$, $f_{yy}$ and $f_{xy}$ for $f(x, y) = e^{2x}\sin y$.
- $f_x = 2e^{2x}\sin y$, so $f_{xx} = 4e^{2x}\sin y$ and $f_{xy} = 2e^{2x}\cos y$.
- $f_y = e^{2x}\cos y$, so $f_{yy} = -e^{2x}\sin y$ (and $f_{yx} = 2e^{2x}\cos y$ again).
Answer: ${}4e^{2x}\sin y$, $-e^{2x}\sin y$, ${}2e^{2x}\cos y$
Worked example: A consistency test
Could $f_x = 2xy + y$ and $f_y = x^2 + 3y^2$ belong to one smooth function?
- Clairaut requires $(f_x)_y = (f_y)_x$.
- $(f_x)_y = 2x + 1$, but $(f_y)_x = 2x$.
- They differ, so no smooth $f$ has these partials.
Answer: no
Common mistake
Reading the order backwards. $f_{xy}$ means $x$ first, then $y$; $\frac{\partial^2 f}{\partial y\,\partial x}$ means the same thing (the variable nearest $f$ goes first). For smooth functions the order does not change the answer, which makes this a perfect check on your algebra: compute both mixed partials, and if they differ, look for a mistake.