Given a fixed value of x, you can find one corresponding y at which the expression is zero as follows
x=3;
f=@(y) x.^2+y.^3-10;
y_zero = fzero(f, y_guess);
Although, when f() is a polynomial, as in your example, it would be more efficient to use ROOTS instead of FZERO. Also, ROOTS will find you all solutions y for that x, instead of just one.
Similarly for given y, you can find one associated x as follows
y=1;
f=@(x) x.^2+y.^3-10;
x_zero = fzero(f, x_guess);
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