No. Conceptually, the reason is that $j'(z)$ is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight $2$, so it cannot be expressed in terms of $j$ (weakly holomorphic modular form of weight $0$) and $z$ (not anywhere near being a modular form).
Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$.
Suppose that the $j$ invariant did satisfy a differential equation of your form. Then we'd have $f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions $z$ and $j(z)$ are algebraically independent (this is just saying that $j(z)$ is a transcendental function). Hence the underlying two-variable rational function $f(x, y)$ satisfies $f(x, y) = f(x, y+1)$. This then easily implies that $f(x, y)$ must be independent of $y$, e.g. $f(x, y) = g(x)$ for some rational function $g$.
So our original differential equation must actually take the form $j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight $0$, 2$while the right hand side has weight 0, and a nonzero modular form has a unique weight, so this is impossible. 1 No. Conceptually, the reason is that$j'(z)$is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight$2$, so it cannot be expressed in terms of$j$(weakly holomorphic modular form of weight$0$) and$z$(not anywhere near being a modular form). For a rigorous proof: Note that$j(z+1) = j(z)$, so$j'(z+1) = j'(z)$. Suppose that the$j$invariant did satisfy a differential equation of your form. Then we'd have$f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions$z$and$j(z)$are algebraically independent (this is just saying that$j(z)$is a transcendental function). Hence the underlying two-variable rational function$f(x, y)$satisfies$f(x, y) = f(x, y+1)$. This then easily implies that$f(x, y)$must be independent of$y$, e.g.$f(x, y) = g(x)$for some rational function$g$. So our original differential equation must actually take the form$j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight$0\$, and a nonzero modular form has a unique weight, so this is impossible.