Cross product partial derivative?

There are actually a whole bunch of partial derivatives there. The curl is a complicated expression involving partials with respect to x, y and z.

The fact that you can do this has to do with a property of functions that the order of differentiation doesn't matter. @H/[@y@x) is the same as @H/(@x@y). I'm using @ for the partial derivative sign. That is, @[@H/@y]/@x is the same as @[@H/@x]/@y. You get that second derivative by differentiating by one variable, then the other, in either order.

So let's look at one component of that expression, say the x component. The x component of curl (-@H/@t) is @/@y[-@Hz/@t] - @/@z[-@Hy/dt]. By the above property of derivatives, we can interchange the order of each of those differentiations:

= @/@t[-@Hz/@y] - @/@t [ -@Hy/@z]

= -@/@t[ @Hz/@y - @Hy/@z ]

and the thing in brackets is the x component of curl H.

Same thing for the y and z components. Everywhere the left hand side has the x, y or z derivative of some component of @H/@t, you can swap the order of the derivatives and write it as the time derivative of the x, y, or z derivative of that component of H.

It's good to be skeptical. It's a good idea when you see these vector identities to confirm them at least once component by component to see how they work and why they work. That will also give you practice in understanding these operators.

(Technical point: the components of Hx, Hy, Hz and these particular first and second derivatives have to be continuous in order to be able to swap order of differentiation. This is obviously physics, and most of the time in physics you assume your functions are well behaved).